Korean, Edit

Inequality Proof Problems [151-200]

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Restructured the IneqMath training data.

P151. Let $a, b, c$ be positive real numbers such that $ab + bc + ca \leq 3abc$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(\sqrt{\frac{a^2+b^2}{a+b}}+\sqrt{\frac{b^2+c^2}{b+c}}+\sqrt{\frac{c^2+a^2}{c+a}}+3 \leq C(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}).\)

P152. Let $a, b, c > 0$. Determine the largest constant $C$ such that the following inequality holds for all positive $a, b, c$:

P153. Let $x, y, z > 0$. Determine the minimal constant $C$ such that the following inequality holds for all $x, y, z$:

P154. Let $a, b, c$ be positive real numbers such that $abc = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$: \(\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2} \geq C.\)

P155. Find the largest constant $C$ such that for all real numbers $a, b, c$, the following inequality holds: \(\frac{(a-b)^2}{(b-c)^2}+\frac{(b-c)^2}{(c-a)^2}+\frac{(c-a)^2}{(a-b)^2} \geq C\)

P156. Given positive real numbers $a, b, c$ satisfying $abc = 1$, determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:

P157. In an isosceles triangle $\triangle ABC$ with $\overline{AB} = \overline{AC}$, the two tangents to the circumcircle of the triangle at $A$ and $B$ intersect at $D$. Determine the minimal constant $C$ such that the following inequality holds for all such triangles: \(\angle DCB \leq C.\)

P158. Let $a, b, c \in \mathbb{R}$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$: \((a+b)^4+(b+c)^4+(c+a)^4 \geq C(a^4+b^4+c^4)\)

P159. Let $n$ be an odd integer greater than 3 and $x_1, x_2, \ldots, x_n \in \mathbb{R}$. Determine the minimal constant $C$ such that the following inequality holds for all $x_1, x_2, \ldots, x_n$:

P160. Let $a, b, c > 0$ such that $abc = 1$. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$:

P161. Let $x \geq y \geq z \geq 0$. Find the largest constant $C$ such that the following inequality holds for all $x, y, z$: \(\frac{x^2 y}{z}+\frac{y^2 z}{x}+\frac{z^2 x}{y} \geq C(x^2+y^2+z^2)\)

P162. Let $0 \leq x \leq 1$ and $n \in \mathbb{N}$. Find the smallest constant $C$ such that the following inequality holds for all $x$ and $n$: \(\left(1-x+\frac{x^2}{2}\right)^n-(1-x)^n \leq Cx.\)

P163. For any acute triangle ( \triangle ABC ) with sides ( a, b, c ), circumradius ( R ), and inradius ( r ), find the largest constant ( C ) such that the following inequality holds: \(a + b + c \geq CR+2r.\)

P164. Let $P$ be a point inside the triangle $ABC$, and let $l = PA$, $m = PB$, and $n = PC$. Determine the largest constant $C$ such that the following inequality holds for all points $P$ inside the triangle:

P165. Let $ABC$ be an acute triangle with circumcenter $O$ and circumradius $R$. If $AO$ intersects the circumcircle of $OBC$ at $D$, $BO$ intersects the circumcircle of $OCA$ at $E$, and $CO$ intersects the circumcircle of $OAB$ at $F$, find the largest constant $C$ such that the following inequality holds for all such triangles: \(O D \cdot O E \cdot O F \geq C R^3.\)

P166. Let $a, b, c$ be the sides of a triangle. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the triangle inequality: \(\sqrt{C a b c+(a+b-c)(b+c-a)(a+c-b)}+\sum_{cyc} a \sqrt{a} \geq \sum_{cyc} a(\sqrt{b}+\sqrt{c}).\)

P167. Let $a_1, \ldots, a_n, b_1, \ldots, b_n > 0$. Find the smallest constant $C$ such that the following inequality holds for all positive $a_i$ and $b_i$: \(\sum_{i=1}^n \frac{a_i b_i}{a_i+b_i} \leq C \cdot \frac{\left(\sum_{i=1}^n a_i\right)\left(\sum_{i=1}^n b_i\right)}{\sum_{i=1}^n\left(a_i+b_i\right)}\)

P168. Let $a, b, c > 0$ such that $a+b+c=3$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:

P169. Let $a, b, c \in \left[\frac{1}{3}, 3\right]$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ in the given range: \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \geq C.\)

P170. Let $a, b, c$ be positive real numbers such that $abc = 1$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c$:

P171. Let $a, b, c$ be the side lengths of a triangle. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the triangle inequalities: \(\frac{a}{3 a-b+c}+\frac{b}{3 b-c+a}+\frac{c}{3 c-a+b} \geq C.\)

P172. Let $a, b, c, d$ be positive real numbers such that $a+b+c+d=4$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c, d$ satisfying the given constraint: \(\frac{a}{1+b^2 c}+\frac{b}{1+c^2 d}+\frac{c}{1+d^2 a}+\frac{d}{1+a^2 b} \geq C\)

P173. Let $a, b, c > 0$ and $a^2 + 2b + c^2 = 1$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$: \(2(a^2 + b^2 + c^2) - (ab + bc + ca) \geq C.\)

P174. Let $x, y, z > 0$. Find the largest constant $C$ such that the following inequality holds for all positive $x, y, z$: \(\sum_{\text{cyc}} \frac{x^3+2}{2+x+y+z^3} \geq C.\)

P175. Let $x, y, z$ be real numbers such that $x+y+z=xyz$. Find the largest constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given constraint: \(\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+z^2} \geq C.\)

P176. Find the largest constant $C$ such that for all real numbers $x$, the following inequality holds: \(x^2+\frac{1}{6(1+|x-2|)} \geq C\)

P177. Let $x_1, x_2, \ldots, x_n$ be positive real numbers such that $x_1 + x_2 + \cdots + x_n = 1$. Find the smallest constant $C$ such that the following inequality holds for all $x_1, x_2, \ldots, x_n$:

where $x_0 = x_n$ and $x_{n+1} = x_1$.

P178. Let $a, b, c$ be positive real numbers such that $a+b+c=ab+bc+ca$ and $n \geq 3$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ and $n$:

P179. Let $a, b, c \geq 0$ and $a^2 + b + c^2 = 1$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:

P180. Let $a, b, c$ be positive real numbers. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{32}{9}\left(\frac{c}{a+b}\right)^2 \geqslant C.\)

P181. There are $2022^2$ integers $a_{i j}(1 \leq i, j \leq 2022)$, every $a_{i j} \in[-2022,2022]$. Find the largest constant $C$ such that the following inequality holds for all $a_{i j}$:

P182. Let $x, y, z$ be complex numbers such that $ㅣx+2y+3zㅣ = 1$. Determine the largest constant $C$ such that the following inequality holds for all $x, y, z$:

P183. Let $a, b, c$ be positive real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: \(\frac{a}{b+C c}+\frac{b}{c+C a}+\frac{c}{a+C b} \geq 1.\)

P184. Let $a, b, c$ be positive real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: \(\frac{a}{\sqrt{b^2+b c+c^2}}+\frac{b}{\sqrt{a^2+a c+c^2}}+\frac{c}{\sqrt{a^2+a b+b^2}} \geq C.\)

P185. Let $a_1, a_2, \ldots, a_n$ be non-negative real numbers such that $\prod a_1 = \lambda^n$. Determine the minimal constant $C$ such that the following inequality holds for all $a_1, a_2, \ldots, a_n$:

P186. Let $x, y, z$ be positive numbers such that $xy + yz + zx = 1$. Find the largest constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given constraint: \(\frac{x^3}{1+9y^2zx} + \frac{y^3}{1+9z^2xy} + \frac{z^3}{1+9x^2yz} \geq C(x+y+z)^3\)

P187. Let $a, b, c \in \mathbb{R}^{+}$. Consider the following inequality: \(\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \quad () \quad \frac{c+2a}{c+b} + \frac{a+2b}{a+c} + \frac{b+2c}{b+a}\)

Determine the correct inequality relation to fill in the blank.

Options:

(A) $\leq$

(B) $\geq$

(C) $=$

(D) $<$

(E) $>$

(F) None of the above

P188. Let $x, y, z \in \mathbb{R}_{+}$. Consider the following inequality: \(\sqrt{x(y+1)} + \sqrt{y(z+1)} + \sqrt{z(x+1)} \quad () \quad 2 \sqrt{(x+1)(y+1)(z+1)}.\)

Determine the correct inequality relation to fill in the blank.

Options:

(A) $\leq$

(B) $\geq$

(C) $=$

(D) $<$

(E) $>$

(F) None of the above

P189. Let $a, b, c > 0$ be positive real numbers such that $a + b + c = 1$. Consider the following inequality: \(\frac{1+a+b}{2+c} + \frac{1+b+c}{2+a} + \frac{1+c+a}{2+b} \quad () \quad 2.\)

Determine the correct inequality relation to fill in the blank.

Options:

(A) $\leq$

(B) $\geq$

(C) $=$

(D) $<$

(E) $>$

(F) None of the above

P190. Let $a, b, c > 0$ be positive real numbers such that $a + b + c = abc$. Consider the following inequality: \(\frac{1}{\sqrt{a^2+1}} + \frac{1}{\sqrt{b^2+1}} + \frac{1}{\sqrt{c^2+1}} \quad () \quad \frac{5}{3}.\)

Determine the correct inequality relation to fill in the blank.

Options:

(A) $\leq$

(B) $\geq$

(C) $=$

(D) $<$

(E) $>$

(F) None of the above

P191. Let $a, b, c$ be positive real numbers such that $a + b + c \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(a + b + c \geq \frac{C}{a+b+c} + \frac{2}{abc}.\)

S191. $C = 3$

By $A M \geq H M$ we get\n\(\na+b+c \\geq \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\geq \\frac{9}{a+b+c}\n\)\ni.e.\n\(\n\\begin{equation*}\n\\frac{a+b+c}{3} \\geq \\frac{3}{a+b+c} \\tag{1}\n\\end{equation*}\n\)\n\nWe will prove that\n\(\n\\begin{equation*}\n\\frac{2(a+b+c)}{3} \\geq \\frac{2}{a b c} \\tag{2}\n\\end{equation*}\n\)\ni.e.\n\(\na+b+c \\geq \\frac{3}{a b c}\n\)\n\nUsing the well-known inequality $(x y+y z+z x)^{2} \geq 3(x y+y z+z x)$ we obtain\n\(\n(a+b+c)^{2} \\geq\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)^{2} \\geq 3\\left(\\frac{1}{a b}+\\frac{1}{b c}+\\frac{1}{c a}\\right)=3 \\frac{a+b+c}{a b c}\n\)\ni.e.\n\(\na+b+c \\geq \\frac{3}{a b c}\n\)\n\nAfter adding (1) and (2) we get the required inequality.\n\nEquality holds when $a = b = c = 1$, since then $a + b + c = 3$, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 3$, and $abc = 1$, so the original constraint is satisfied with equality, and the inequality becomes $3 \geq \frac{3}{3} + 2 = 3$. This gives the maximum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = 3$.

P192. Let $a, b, c, x, y, z$ be positive real numbers such that $x + y + z = 1$. Determine the maximal constant $C$ such that the following inequality holds for all $a, b, c, x, y, z$:

S192. $C = 2$

The inequality being homogeneous in $a, b, c$ we can assume that $a+b+c=1$. We apply this time the AM-GM Inequality and we find that\n$a x+b y+c z+2 \sqrt{(x y+y z+z x)(a b+b c+c a)} \leq a x+b y+c z+x y+y z+z x+a b+b c+c a$.\nConsequently,\n$x y+y z+z x+a b+b c+c a=\frac{1-x^{2}-y^{2}-z^{2}}{2}+\frac{1-a^{2}-b^{2}-c^{2}}{2} \leq 1-a x-b y-c z$,\nthe last one being equivalent to $(x-a)^{2}+(y-b)^{2}+(z-c)^{2} \geq 0$.\n\nEquality holds when $x = a$, $y = b$, $z = c$, that is, when the vectors $(x, y, z)$ and $(a, b, c)$ are equal. In this case, the inequality becomes an equality, and $C = 2$ gives the maximal value for which the inequality always holds.\n\nTherefore, the answer is $C = 2$.”, “We will use the Cauchy-Schwarz Inequality twice. First, we can write $a x+b y+$ $c z \leq \sqrt{a^{2}+b^{2}+c^{2}} \cdot \sqrt{x^{2}+y^{2}+z^{2}}$ and then we apply again the Cauchy-Schwarz Inequality to obtain:\n\n\(\n\\begin{aligned}\na x+b y+c z & +2 \\sqrt{(x y+y z+z x)(a b+b c+c a)} \\leq \\\\\n& \\leq \\sqrt{\\sum a^{2}} \\cdot \\sqrt{\\sum x^{2}}+\\sqrt{2 \\sum a b} \\cdot \\sqrt{2 \\sum x y} \\leq \\\\\n& \\leq \\sqrt{\\sum x^{2}+2 \\sum x y} \\cdot \\sqrt{\\sum a^{2}+2 \\sum a b}=\\sum a\n\\end{aligned}\n\)\n\nEquality in the above inequalities holds when $a = b = c$ and $x = y = z$, that is, when all variables are equal. In this case, the inequality becomes an equality, and $C = 2$ gives the maximal value for which the inequality always holds.\n\nTherefore, the answer is $C = 2$.

P193. Let $a, b, c$ be positive real numbers such that $ab + bc + ca = 3$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(\left(a^{3}-a+5\right)\left(b^{5}-b^{3}+5\right)\left(c^{7}-c^{5}+5\right) \geq C\)

S193. $C = 125$

For any real number $x$, the numbers $x-1, x^{2}-1, x^{3}-1$ and $x^{5}-1$ are of the same sign.\n\nTherefore\n\(\n(x-1)\\left(x^{2}-1\\right) \\geq 0, \\quad\\left(x^{2}-1\\right)\\left(x^{3}-1\\right) \\geq 0 \\quad \\text { and } \\quad\\left(x^{2}-1\\right)\\left(x^{5}-1\\right) \\geq 0\n\)\ni.e.\n\(\n\\begin{aligned}\na^{3}-a^{2}-a+1 & \\geq 0 \\\\\nb^{5}-b^{3}-b^{2}+1 & \\geq 0 \\\\\nc^{7}-c^{5}-c^{2}+1 & \\geq 0\n\\end{aligned}\n\)\n\nSo it follows that\n\(\na^{3}-a+5 \\geq a^{2}+4, \\quad b^{5}-b^{3}+5 \\geq b^{2}+4 \\quad \\text { and } \\quad c^{7}-c^{5}+5 \\geq c^{2}+4\n\)\n\nMultiplying these inequalities gives us\n\(\n\\begin{equation*}\n\\left(a^{3}-a+5\\right)\\left(b^{5}-b^{3}+5\\right)\\left(c^{7}-c^{5}+5\\right) \\geq\\left(a^{2}+4\\right)\\left(b^{2}+4\\right)\\left(c^{2}+4\\right) \\tag{1}\n\\end{equation*}\n\)\n\nWe’ll prove that\n\(\n\\begin{equation*}\n\\left(a^{2}+4\\right)\\left(b^{2}+4\\right)\\left(c^{2}+4\\right) \\geq 25(a b+b c+c a+2) \\tag{2}\n\\end{equation*}\n\)\n\nWe have\n\(\n\\begin{aligned}\n& \\left(a^{2}+4\\right)\\left(b^{2}+4\\right)\\left(c^{2}+4\\right) \\\\\n& \\quad=a^{2} b^{2} c^{2}+4\\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\\right)+16\\left(a^{2}+b^{2}+c^{2}\\right)+64\n\\end{aligned}\n\)\n\(\n\\begin{align*}\n= & a^{2} b^{2} c^{2}+\\left(a^{2}+b^{2}+c^{2}\\right)+2+4\\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+3\\right) \\\\\n& +15\\left(a^{2}+b^{2}+c^{2}\\right)+50 \\tag{3}\n\\end{align*}\n\)\n\nBy the obvious inequalities\n$(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0 \quad$ and $\quad(a b-1)^{2}+(b c-1)^{2}+(c a-1)^{2} \geq 0$\nwe obtain\n\(\n\\begin{align*}\na^{2}+b^{2}+c^{2} & \\geq a b+b c+c a \\tag{4}\\\\\na^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+3 & \\geq 2(a b+b c+c a) \\tag{5}\n\\end{align*}\n\)\n\nWe’ll prove that\n\(\n\\begin{equation*}\na^{2} b^{2} c^{2}+\\left(a^{2}+b^{2}+c^{2}\\right)+2 \\geq 2(a b+b c+c a) \\tag{6}\n\\end{equation*}\n\)\n\nLemma Let $x, y, z>0$. Then\n\(\n3 x y z+x^{3}+y^{3}+z^{3} \\geq 2\\left((x y)^{3 / 2}+(y z)^{3 / 2}+(z x)^{3 / 2}\\right)\n\)\n\nProof By Schur’s inequality and $A M \geq G M$ we have\n\(\n\\begin{aligned}\nx^{3}+y^{3}+z^{3}+3 x y z & \\geq\\left(x^{2} y+y^{2} x\\right)+\\left(z^{2} y+y^{2} z\\right)+\\left(x^{2} z+z^{2} x\\right) \\\\\n& \\geq 2\\left((x y)^{3 / 2}+(y z)^{3 / 2}+(z x)^{3 / 2}\\right)\n\\end{aligned}\n\)\n\nBy Lemma for $x=a^{2 / 3}, y=b^{2 / 3}, z=c^{2 / 3}$ we deduce\n\(\n3(a b c)^{2 / 3}+a^{2}+b^{2}+c^{2} \\geq 2(a b+b c+c a)\n\)\n\nTherefore it suffices to prove that\n\(\na^{2} b^{2} c^{2}+2 \\geq 3(a b c)^{2 / 3}\n\)\nwhich follows immediately by $A M \geq G M$.\nThus we have proved inequality (6).\nNow by (3), (4), (5) and (6) we obtain inequality (2).\nFinally by (1), (2) and since $a b+b c+c a=3$ we obtain the required inequality. Equality occurs if and only if $a=b=c=1$.\n\nEquality holds when $a = b = c = 1$, which satisfies $ab + bc + ca = 3$. In this case, the expression evaluates to $(1^3 - 1 + 5)(1^5 - 1^3 + 5)(1^7 - 1^5 + 5) = (5)(5)(5) = 125$. This gives the minimum value of the expression, so $C = 125$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 125$.

P194. Let $n$ be an integer greater than 2. Find the greatest real number $C_{\text{min}}$ and the least real number $C_{\text{max}}$ such that for any positive real numbers $x_1, x_2, \ldots, x_n$ (with $x_n = x_0$ and $x_{n+1} = x_1$), the following inequality holds: \(C_{\text{min}} \leq \sum_{i=1}^{n} \frac{x_i}{x_{i-1} + 2(n-1)x_i + x_{i+1}} \leq C_{\text{max}}\)

S194. $C_{\text{min}} = \frac{1}{2(n-1)}$ $C_{\text{max}} = \frac{1}{2}$

We will prove that $m_{n}=\frac{1}{2(n-1)}, M_{n}=\frac{1}{2}$. First, let us see that the inequality\n\n\(\n\\sum_{i=1}^{n} \\frac{x_{i}}{x_{i-1}+2(n-1) x_{i}+x_{i+1}} \\geq \\frac{1}{2(n-1)}\n\)\n\nis trivial, because $x_{i-1}+2(n-1) x_{i}+x_{i+1} \leq 2(n-1) \cdot \sum_{k=1}^{n} x_{k}$ for all $i$. This shows that $m_{n} \geq \frac{1}{2(n-1)}$. Taking $x_{i}=x^{i}$, the expression becomes\n\n\(\n\\frac{1}{x+x^{n-1}+2(n-1)}+\\frac{(n-2) x}{1+2(n-1) x+x^{2}}+\\frac{x^{n-1}}{1+2(n-1) x^{n-1}+x^{n-2}}\n\)\n\nand taking the limit when $x$ approaches 0 , we find that $m_{n} \leq \frac{1}{2(n-1)}$ and thus $m_{n}=\frac{1}{2(n-1)}$\n\nNow, we will prove that $M_{n} \geq \frac{1}{2}$. Of course, it suffices to prove that for any $x_{1}, x_{2}, \ldots, x_{n}>0$ we have\n\n\(\n\\sum_{i=1}^{n} \\frac{x_{i}}{x_{i-1}+2(n-1) x_{i}+x_{i+1}} \\leq \\frac{1}{2}\n\)\n\nBut it is clear that\n\n\(\n\\begin{gathered}\n\\sum_{i=1}^{n} \\frac{2 x_{i}}{x_{i-1}+2(n-1) x_{i}+x_{i+1}} \\leq \\sum_{i=1}^{n} \\frac{2 x_{i}}{2 \\sqrt{x_{i-1} \\cdot x_{i+1}}+2(n-1) x_{i}}= \\\\\n=\\sum_{i=1}^{n} \\frac{1}{n-1+\\frac{\\sqrt{x_{i-1} \\cdot x_{i+1}}}{x_{i}}}\n\\end{gathered}\n\)\n\nTaking $\frac{\sqrt{x_{i-1} \cdot x_{i+1}}}{x_{i}}=a_{i}$, we have to prove that if $\prod_{i=1}^{n} a_{i}=1$ then $\sum_{i=1}^{n} \frac{1}{n-1+a_{i}} \leq 1$. But this has already been proved in the problem 84. Thus, $M_{n} \geq \frac{1}{2}$ and because for $x_{1}=x_{2}=\cdots=x_{n}$ we have equality, we deduce that $M_{n}=\frac{1}{2}$, which solves the problem.\n\nEquality in the lower bound $C_{\text{min}} = \frac{1}{2(n-1)}$ is achieved in the limit as one variable tends to zero and the others are fixed, for example, by taking $x_1 = x_2 = \cdots = x_{n-1} = 1$ and $x_n \to 0$. Equality in the upper bound $C_{\text{max}} = \frac{1}{2}$ is achieved when all variables are equal, i.e., $x_1 = x_2 = \cdots = x_n$. Thus, $C_{\text{min}}$ is the minimum value and $C_{\text{max}}$ is the maximum value of the given sum.\n\nTherefore, the answer is \(C_{\\text{min}} = \\frac{1}{2(n-1)}\)\n\(C_{\\text{max}} = \\frac{1}{2}\).

P195. Let $x_{1}, x_{2}, \ldots, x_{n}$ be non-negative real numbers such that $x_{1} + x_{2} + \cdots + x_{n} \leq \frac{1}{2}$. Find the largest constant $C$ such that the following inequality holds for all $x_{1}, x_{2}, \ldots, x_{n}$ satisfying the given constraint: \(\left(1 - x_{1}\right)\left(1 - x_{2}\right) \cdots \left(1 - x_{n}\right) \geq C\)

S195. $C = \frac{1}{2}$

From $x_{1}+x_{2}+\cdots+x_{n} \leq \frac{1}{2}$ and the fact that $x_{1}, x_{2}, \ldots, x_{n}$ are nonnegative we deduce that\n\(\n0 \\leq x_{i} \\leq \\frac{1}{2}<1, \\quad \\text { i.e. } \\quad-x_{i}>-1, \\quad \\text { for all } i=1,2, \\ldots, n\n\)\nand it’s clear that all $-x_{i}$ are of the same sign.\nWe obtain\n\(\n\\begin{aligned}\n\\left(1-x_{1}\\right)\\left(1-x_{2}\\right) \\cdots\\left(1-x_{n}\\right) & =\\left(1+\\left(-x_{1}\\right)\\right)\\left(1+\\left(-x_{2}\\right)\\right) \\cdots\\left(1+\\left(-x_{n}\\right)\\right) \\\\\n& \\geq 1+\\left(-x_{1}-x_{2}-\\cdots-x_{n}\\right) \\\\\n& =1-\\left(x_{1}+x_{2}+\\cdots+x_{n}\\right) \\geq 1-\\frac{1}{2}=\\frac{1}{2}\n\\end{aligned}\n\)\nEquality holds when $x_1 = \frac{1}{2}$ and $x_2 = x_3 = \cdots = x_n = 0$. This gives the minimum value of the product, so $C = \frac{1}{2}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{1}{2}$.

P196. Let $a, b, c \in (-3, 3)$ such that $\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c}=\frac{1}{3-a}+\frac{1}{3-b}+\frac{1}{3-c}$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given condition: \(\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c} \geq C\)

S196. $C = 1$

By the inequality $A M \geq H M$ we have\n\(\n\\begin{equation*}\n((3+a)+(3+b)+(3+c))\\left(\\frac{1}{3+a}+\\frac{1}{3+b}+\\frac{1}{3+c}\\right) \\geq 9 \\tag{1}\n\\end{equation*}\n\)\nand\n\(\n\\begin{align*}\n& ((3-a)+(3-b)+(3-c))\\left(\\frac{1}{3-a}+\\frac{1}{3-b}+\\frac{1}{3-c}\\right) \\geq 9 \\\\\n& \\quad \\Leftrightarrow \\quad((3-a)+(3-b)+(3-c))\\left(\\frac{1}{3+a}+\\frac{1}{3+b}+\\frac{1}{3+c}\\right) \\geq 9 \\tag{2}\n\\end{align*}\n\)\n\nAfter adding (1) and (2) we obtain\n\(\n18\\left(\\frac{1}{3+a}+\\frac{1}{3+b}+\\frac{1}{3+c}\\right) \\geq 18, \\quad \\text { i.e. } \\quad \\frac{1}{3+a}+\\frac{1}{3+b}+\\frac{1}{3+c} \\geq 1\n\)\n\nEquality holds when $a = b = c = 0$, since then $\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c} = 3 \times \frac{1}{3} = 1$. This gives the minimum value of the sum, so $C = 1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P197. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(\frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq C\)

S197. $C = 1$

Applying $A M \geq G M$ gives us\n\(\n\\frac{a^{2}}{a+2 b^{3}}=a-\\frac{2 a b^{3}}{a+2 b^{3}} \\geq a-\\frac{2 a b^{3}}{3 \\sqrt[3]{a b^{4}}}=a-\\frac{2 b a^{2 / 3}}{3}\n\)\n\nAnalogously\n\(\n\\frac{b^{2}}{b+2 c^{3}} \\geq b-\\frac{2 c b^{2 / 3}}{3} \\quad \\text { and } \\quad \\frac{c^{2}}{c+2 a^{3}} \\geq c-\\frac{2 a c^{2 / 3}}{3}\n\)\n\nAdding these three inequalities implies\n\(\n\\frac{a^{2}}{a+2 b^{2}}+\\frac{b^{2}}{b+2 c^{2}}+\\frac{c^{2}}{c+2 a^{2}} \\geq(a+b+c)-\\frac{2}{3}\\left(b a^{2 / 3}+c b^{2 / 3}+a c^{2 / 3}\\right)\n\)\n\nSo it is enough to prove that\n\(\n(a+b+c)-\\frac{2}{3}\\left(b a^{2 / 3}+c b^{2 / 3}+a c^{2 / 3}\\right) \\geq 1\n\)\ni.e.\n\(\n\\begin{equation*}\nb a^{2 / 3}+c b^{2 / 3}+a c^{2 / 3} \\leq 3 \\tag{1}\n\\end{equation*}\n\)\n\nAfter another application of $A M \geq G M$ we get\n\(\n\\begin{aligned}\nb a^{2 / 3}+c b^{2 / 3}+a c^{2 / 3} & \\leq \\frac{b(2 a+1)+c(2 b+1)+a(2 c+1)}{3} \\\\\n& =\\frac{a+b+c+2(a b+b c+c a)}{3} \\\\\n& \\leq \\frac{(a+b+c)+(a+b+c)^{2} / 3}{3}=\\frac{3+2 \\cdot 3^{2} / 3}{3}=3\n\\end{aligned}\n\)\ni.e. we have proved (1), and we are done.\n\nEquality holds when $a = b = c = 1$, since then $a + b + c = 3$ and all terms are equal, so the minimum value of the sum is achieved and $C = 1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P198. Let $t_{a}, t_{b}, t_{c}$ be the lengths of the medians, and $a, b, c$ be the lengths of the sides of a given triangle. Find the smallest constant $C$ such that the following inequality holds for all triangles: \(t_{a} t_{b}+t_{b} t_{c}+t_{c} t_{a} < C(a b+b c+c a)\)

S198. $C = \frac{5}{4}$

We can easily show the inequalities\n\(\nt_{a}<\\frac{b+c}{2}, \\quad t_{b}<\\frac{a+c}{2}, \\quad t_{c}<\\frac{b+a}{2}\n\)\n\nAfter adding these we get\n\(\n\\begin{equation*}\nt_{a}+t_{b}+t_{c}<a+b+c \\tag{1}\n\\end{equation*}\n\)\n\nBy squaring (1) we deduce\n\(\n\\begin{equation*}\nt_{a}^{2}+t_{b}^{2}+t_{c}^{2}+2\\left(t_{a} t_{b}+t_{b} t_{c}+t_{c} t_{a}\\right)<a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \\tag{2}\n\\end{equation*}\n\)\n\nOn the other hand, we have\n\(\nt_{a}^{2}=\\frac{2\\left(b^{2}+c^{2}\\right)-a^{2}}{4}, \\quad t_{b}^{2}=\\frac{2\\left(a^{2}+c^{2}\\right)-b^{2}}{4}, \\quad t_{c}^{2}=\\frac{2\\left(b^{2}+a^{2}\\right)-c^{2}}{4}\n\)\nso\n\(\nt_{a}^{2}+t_{b}^{2}+t_{c}^{2}=\\frac{3}{4}\\left(a^{2}+b^{2}+c^{2}\\right)\n\)\n\nNow using the previous result and (2) we get\n\(\n\\begin{equation*}\nt_{a} t_{b}+t_{b} t_{c}+t_{c} t_{a}<\\frac{1}{8}\\left(a^{2}+b^{2}+c^{2}\\right)+(a b+b c+c a) \\tag{3}\n\\end{equation*}\n\)\n\nAlso we have $a^{2}+b^{2}+c^{2}<2(a b+b c+c a)$, since\n$a^{2}+b^{2}+c^{2}-2(a b+b c+c a)=a(a-b-c)+b(b-a-c)+c(c-a-b)<0$.\nFinally by (3) and the previous inequality we obtain\n\(\nt_{a} t_{b}+t_{b} t_{c}+t_{c} t_{a}<\\frac{5}{4}(a b+b c+c a)\n\)\n\nEquality in the above inequalities is never achieved for any non-degenerate triangle, since all the inequalities used are strict. Thus, the value $C = \frac{5}{4}$ is the smallest constant for which the inequality always holds, but the inequality is always strict and never achieved with equality for any triangle.\n\nTherefore, the answer is $C = \frac{5}{4}$.

P199. Let $a, b, c \in \mathbb{R}$ such that $a + b + c \geq abc$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(a^2 + b^2 + c^2 \geq C \, abc.\)

S199. $C = \sqrt{3}$

We have\n\(\n\\begin{align*}\n\\left(a^{2}+b^{2}+c^{2}\\right)^{2} & =a^{4}+b^{4}+c^{4}+2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2} \\\\\n& =a^{4}+b^{4}+c^{4}+a^{2}\\left(b^{2}+c^{2}\\right)+b^{2}\\left(c^{2}+a^{2}\\right)+c^{2}\\left(a^{2}+b^{2}\\right) \\tag{1}\n\\end{align*}\n\)\n\nBy Exercise 1.7, it follows that\n\(\n\\begin{equation*}\na^{4}+b^{4}+c^{4} \\geq a b c(a+b+c) \\tag{2}\n\\end{equation*}\n\)\n\nAlso\n\(\n\\begin{equation*}\nb^{2}+c^{2} \\geq 2 b c, \\quad c^{2}+a^{2} \\geq 2 c a, \\quad a^{2}+b^{2} \\geq 2 a b \\tag{3}\n\\end{equation*}\n\)\n\nNow by (1), (2) and (3) we deduce\n\(\n\\begin{align*}\n\\left(a^{2}+b^{2}+c^{2}\\right)^{2} & \\geq a b c(a+b+c)+2 a^{2} b c+2 b^{2} a c+2 c^{2} a b \\\\\n& =a b c(a+b+c)+2 a b c(a+b+c)=3 a b c(a+b+c) \\tag{4}\n\\end{align*}\n\)\n\nSince $a+b+c \geq a b c$ in (4) we have\n\(\n\\left(a^{2}+b^{2}+c^{2}\\right)^{2} \\geq 3 a b c(a+b+c) \\geq 3(a b c)^{2}\n\)\ni.e.\n\(\na^{2}+b^{2}+c^{2} \\geq \\sqrt{3} a b c .\n\)\n\nEquality holds when $a = b = c = \sqrt{3}$, since then $a + b + c = 3\sqrt{3}$ and $abc = (\sqrt{3})^3 = 3\sqrt{3}$, so the constraint is satisfied with equality, and $a^2 + b^2 + c^2 = 3 \times 3 = 9 = \sqrt{3} \times 3\sqrt{3} = 9$. This gives the minimum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = \sqrt{3}$.

P200. Let $a, b, c \in \mathbb{R}^{+}$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: \(a^{4}+b^{4}+c^{4} \geq C \cdot a b c (a+b+c)\)

S200. $C = 1$

We have\n\(\n\\begin{aligned}\na^{4} & +b^{4}+c^{4} \\geq a b c(a+b+c) \\\\\n& \\Leftrightarrow \\quad a^{4}+b^{4}+c^{4} \\geq a^{2} b c+b^{2} a c+c^{2} a b \\\\\n& \\Leftrightarrow \\quad \\frac{T[4,0,0]}{2} \\geq \\frac{T[2,1,1]}{2}\n\\end{aligned}\n\)\ni.e.\n\(\nT[4,0,0] \\geq T[2,1,1]\n\)\nwhich is true according to Muirhead’s theorem.\n\nEquality holds when $a = b = c$, since then both sides are equal: $a^4 + b^4 + c^4 = 3a^4$ and $abc(a+b+c) = a^3 \cdot 3a = 3a^4$. This gives the minimum value of the left side, so $C = 1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

Input: 2025.12.08 15:51