Inequality Proof Problems [151-200]
Recommended posting: 【Algebra】 Algebra Index
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 $a, b, c$:
\[\frac{4 a}{a+b}+\frac{4 b}{b+c}+\frac{4 c}{c+a}+\frac{a b^2+b c^2+c a^2+a b c}{a^2 b+b^2 c+c^2 a+a b c} \geq C.\]P153. Let $x, y, z > 0$. Determine the minimal constant $C$ such that the following inequality holds for all $x, y, z$:
\[\frac{x y z}{(1+3 x)(x+8 y)(y+9 z)(z+6)} \leq C.\]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 $a, b, c$:
\[\frac{3}{(1+a)^2}+\frac{3}{(1+b)^2}+\frac{3}{(1+c)^2}+\frac{3}{a+b+c+1} \geq C.\]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 $x_1, x_2, \ldots, x_n$:
\[C (n-1)\max \left(x_1^2, x_2^2, \ldots, x_n^2\right)+\left(x_1+x_2+\ldots+x_n\right)^2 \geq x_1^2+x_2^2+\ldots+x_n^2\]P160. Let $a, b, c > 0$ such that $abc = 1$. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$:
\[\left(\frac{a+b+c}{3}\right)^{C} \geq \frac{a^2+b^2+c^2}{3}\]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:
\[(l m + m n + n l)(l + m + n) \geq C(a^2 l + b^2 m + c^2 n).\]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 $a, b, c$:
\[a b c + \frac{17}{a b + b c + c a} \geq C.\]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 $a, b, c$:
\[\frac{a^2 b^2}{a^7+b^7+a^2 b^2}+\frac{b^2 c^2}{b^7+c^7+b^2 c^2}+\frac{c^2 a^2}{c^7+a^7+c^2 a^2} \leq C.\]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 $x_1, x_2, \ldots, x_n$:
\[\sum_{i=1}^n \frac{\min \left\{x_{i-1}, x_i\right\} \cdot \max \left\{x_i, x_{i+1}\right\}}{x_i} \leq C\]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 $n$:
\[\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2} \geq C+n.\]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 $a, b, c$:
\[\frac{1}{1+a^2} + \frac{6}{5(1+b^2)} + \frac{1}{1+c^2} \geq C.\]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 $a_{i j}$:
\[\prod_{i=1}^{2022}\left(\sum_{j=1}^{2022} a_{i j}^2\right) \geq C \left(\sum_{j=1}^{2022} \prod_{i=1}^{2022} a_{i j}\right)^2\]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 $x, y, z$:
\[x^2 + y^2 + z^2 + \left|x^2 + y^2 + z^2\right| \geq C.\]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 $a_1, a_2, \ldots, a_n$:
\[\prod_{c y c}\left(a_1^2-a_1+1\right) \geq \sqrt{\left(\frac{1+\lambda^4}{C}\right)^n}.\]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 $AM \ge HM$ we get
\[a+b+c \ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9}{a+b+c}.\]i.e.
\[\frac{a+b+c}{3} \ge \frac{3}{a+b+c} \qquad (1)\]We will prove that
\[\frac{2(a+b+c)}{3} \ge \frac{2}{abc} \qquad (2)\]i.e.
\[a+b+c \ge \frac{3}{abc}.\]Using the well-known inequality $(xy+yz+zx)^2 \ge 3(xy+yz+zx)$ we obtain
\[(a+b+c)^2 \ge \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2 \ge 3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right) = 3\frac{a+b+c}{abc},\]i.e.
\[a+b+c \ge \frac{3}{abc}.\]After adding (1) and (2) we get the required inequality.
Equality 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 $3 \ge \frac{3}{3}+2 = 3$. This gives the maximum value of $C$ for which the inequality always holds.
Therefore, 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 $a, b, c, x, y, z$:
\[a x + b y + c z + C \sqrt{(x y + y z + z x)(a b + b c + c a)} \leq a+b+c.\]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:
\[\begin{aligned} ax+by+cz + 2\sqrt{(xy+yz+zx)(ab+bc+ca)} &\le \sqrt{\sum a^{2}}\sqrt{\sum x^{2}} + \sqrt{2\sum ab}\sqrt{2\sum xy} \\ &\le \sqrt{\sum x^{2}+2\sum xy}\,\sqrt{\sum a^{2}+2\sum ab} \\ &= \sum a. \end{aligned}\]Equality 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.
Therefore,
\[(x-1)(x^{2}-1)\ge 0,\quad (x^{2}-1)(x^{3}-1)\ge 0,\quad (x^{2}-1)(x^{5}-1)\ge 0.\]i.e.
\[\begin{aligned} a^{3}-a^{2}-a+1 &\ge 0,\\ b^{5}-b^{3}-b^{2}+1 &\ge 0,\\ c^{7}-c^{5}-c^{2}+1 &\ge 0. \end{aligned}\]So it follows that
\[a^{3}-a+5 \ge a^{2}+4,\quad b^{5}-b^{3}+5 \ge b^{2}+4,\quad c^{7}-c^{5}+5 \ge c^{2}+4.\]Multiplying these inequalities gives
\[(a^{3}-a+5)(b^{5}-b^{3}+5)(c^{7}-c^{5}+5)\ge (a^{2}+4)(b^{2}+4)(c^{2}+4). \qquad (1)\]We will prove that
\[(a^{2}+4)(b^{2}+4)(c^{2}+4)\ge 25(ab+bc+ca+2). \qquad (2)\]We have
\[\begin{aligned} (a^{2}+4)(b^{2}+4)(c^{2}+4) &=a^{2}b^{2}c^{2}+4(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})+16(a^{2}+b^{2}+c^{2})+64\\ &=a^{2}b^{2}c^{2}+(a^{2}+b^{2}+c^{2})+2 +4(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+3)\\ &\quad +15(a^{2}+b^{2}+c^{2})+50. \qquad (3) \end{aligned}\]By the obvious inequalities
\[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\ge 0,\qquad (ab-1)^{2}+(bc-1)^{2}+(ca-1)^{2}\ge 0,\]we obtain
\[a^{2}+b^{2}+c^{2}\ge ab+bc+ca, \qquad (4)\]and
\[a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+3\ge 2(ab+bc+ca). \qquad (5)\]We will prove that
\[a^{2}b^{2}c^{2}+(a^{2}+b^{2}+c^{2})+2 \ge 2(ab+bc+ca). \qquad (6)\]Lemma. Let $x,y,z>0$. Then
\[3xyz+x^{3}+y^{3}+z^{3} \ge 2\big((xy)^{3/2}+(yz)^{3/2}+(zx)^{3/2}\big).\]Proof. By Schur’s inequality and $AM\ge GM$ we have
\[\begin{aligned} x^{3}+y^{3}+z^{3}+3xyz &\ge (x^{2}y+y^{2}x)+(z^{2}y+y^{2}z)+(x^{2}z+z^{2}x)\\ &\ge 2\big((xy)^{3/2}+(yz)^{3/2}+(zx)^{3/2}\big). \end{aligned}\]By the lemma for $x=a^{2/3}$, $y=b^{2/3}$, $z=c^{2/3}$ we deduce
\[3(abc)^{2/3}+a^{2}+b^{2}+c^{2}\ge 2(ab+bc+ca).\]Therefore it suffices to prove that
\[a^{2}b^{2}c^{2}+2 \ge 3(abc)^{2/3},\]which follows immediately by $AM\ge GM$. Thus we have proved (6).
Now by (3), (4), (5), and (6) we obtain (2). Finally, by (1), (2), and since $ab+bc+ca=3$ we obtain the required inequality.
Equality holds if and only if $a=b=c=1$. In this case,
\[(1^{3}-1+5)(1^{5}-1^{3}+5)(1^{7}-1^{5}+5)=5\cdot 5\cdot 5=125,\]so the largest constant 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
\[\sum_{i=1}^{n}\frac{x_i}{x_{i-1}+2(n-1)x_i+x_{i+1}} \ge \frac{1}{2(n-1)}\]is trivial, because
\[x_{i-1}+2(n-1)x_i+x_{i+1}\le 2(n-1)\sum_{k=1}^{n}x_k \quad \text{for all } i.\]This shows that $(m_n\ge \frac{1}{2(n-1)})$. Taking $(x_i=x^i)$, the expression becomes
\[\frac{1}{x+x^{n-1}+2(n-1)} +\frac{(n-2)x}{1+2(n-1)x+x^{2}} +\frac{x^{n-1}}{1+2(n-1)x^{n-1}+x^{n-2}}.\]Taking the limit as (x\to 0), we find that (m_n\le \frac{1}{2(n-1)}), and thus
\[m_n=\frac{1}{2(n-1)}.\]Now, we will prove that $(M_n\le \frac{1}{2})$. Of course, it suffices to prove that for any $(x_1,x_2,\ldots,x_n>0)$ we have
\[\sum_{i=1}^{n}\frac{x_i}{x_{i-1}+2(n-1)x_i+x_{i+1}} \le \frac{1}{2}.\]But it is clear that
\[\begin{aligned} \sum_{i=1}^{n}\frac{2x_i}{x_{i-1}+2(n-1)x_i+x_{i+1}} &\le \sum_{i=1}^{n}\frac{2x_i}{2\sqrt{x_{i-1}x_{i+1}}+2(n-1)x_i}\\ &= \sum_{i=1}^{n}\frac{1}{\,n-1+\frac{\sqrt{x_{i-1}x_{i+1}}}{x_i}\,}. \end{aligned}\]Let $(a_i=\frac{\sqrt{x_{i-1}x_{i+1}}}{x_i})$. Then $(\prod_{i=1}^{n}a_i=1)$, and we have to prove that
\[\sum_{i=1}^{n}\frac{1}{n-1+a_i}\le 1.\]But this has already been proved in Problem 84. Hence
\[\sum_{i=1}^{n}\frac{2x_i}{x_{i-1}+2(n-1)x_i+x_{i+1}}\le 1,\]so
\[\sum_{i=1}^{n}\frac{x_i}{x_{i-1}+2(n-1)x_i+x_{i+1}}\le \frac{1}{2}.\]Therefore $(M_n\le \frac{1}{2})$, and because for $(x_1=x_2=\cdots=x_n)$ we have equality, we deduce that
\[M_n=\frac{1}{2}.\]Equality in the lower bound $(C_{\min}=\frac{1}{2(n-1)})$ 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_{\max}=\frac{1}{2}) is achieved when all variables are equal, i.e. (x_1=x_2=\cdots=x_n).
Therefore, the answer is
\[C_{\min}=\frac{1}{2(n-1)},\qquad C_{\max}=\frac{1}{2}.\]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 $(x_{1}+x_{2}+\cdots+x_{n}\le \frac12)$ and the fact that $(x_{1},x_{2},\ldots,x_{n})$ are nonnegative, we deduce that
\[0\le x_i \le \frac12 < 1, \quad\text{i.e.}\quad -x_i > -1, \quad\text{for all } i=1,2,\ldots,n.\]Hence all $(-x_i)$ are of the same sign. We obtain
\[\begin{aligned} (1-x_1)(1-x_2)\cdots(1-x_n) &=(1+(-x_1))(1+(-x_2))\cdots(1+(-x_n))\\ &\ge 1+(-x_1-x_2-\cdots-x_n)\\ &=1-(x_1+x_2+\cdots+x_n)\\ &\ge 1-\frac12=\frac12. \end{aligned}\]Equality holds when $(x_1=\frac12)$ and $(x_2=x_3=\cdots=x_n=0)$. Thus the minimum value of the product is $(\frac12)$, so the largest constant $(C)$ for which the inequality always holds is
\[C=\frac12.\]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 $(AM \ge HM)$ we have
\[\big((3+a)+(3+b)+(3+c)\big)\left(\frac1{3+a}+\frac1{3+b}+\frac1{3+c}\right)\ge 9. \qquad (1)\]Also,
\[\big((3-a)+(3-b)+(3-c)\big)\left(\frac1{3-a}+\frac1{3-b}+\frac1{3-c}\right)\ge 9. \qquad (2)\]After adding (1) and (2) we obtain
\[\Big((3+a)+(3+b)+(3+c)+(3-a)+(3-b)+(3-c)\Big) \left(\frac1{3+a}+\frac1{3+b}+\frac1{3+c}\right)\ge 18,\]hence
\[18\left(\frac1{3+a}+\frac1{3+b}+\frac1{3+c}\right)\ge 18, \quad\text{i.e.}\quad \frac1{3+a}+\frac1{3+b}+\frac1{3+c}\ge 1.\]Equality holds when $(a=b=c=0)$, since then
\[\frac1{3+a}+\frac1{3+b}+\frac1{3+c}=3\cdot\frac13=1.\]Therefore, 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 $(AM \ge GM)$ gives
\[\frac{a^{2}}{a+2b^{3}} = a-\frac{2ab^{3}}{a+2b^{3}} \ge a-\frac{2ab^{3}}{3\sqrt[3]{ab^{4}}} = a-\frac{2}{3}ba^{2/3}.\]Analogously,
\[\frac{b^{2}}{b+2c^{3}} \ge b-\frac{2}{3}cb^{2/3}, \qquad \frac{c^{2}}{c+2a^{3}} \ge c-\frac{2}{3}ac^{2/3}.\]Adding these three inequalities implies
\[\frac{a^{2}}{a+2b^{3}}+\frac{b^{2}}{b+2c^{3}}+\frac{c^{2}}{c+2a^{3}} \ge (a+b+c)-\frac{2}{3}\big(ba^{2/3}+cb^{2/3}+ac^{2/3}\big).\]So it is enough to prove that
\[(a+b+c)-\frac{2}{3}\big(ba^{2/3}+cb^{2/3}+ac^{2/3}\big)\ge 1,\]i.e.
\[ba^{2/3}+cb^{2/3}+ac^{2/3}\le 3. \qquad (1)\]After another application of $(AM \ge GM)$ we get
\[\begin{aligned} ba^{2/3}+cb^{2/3}+ac^{2/3} &\le \frac{b(2a+1)+c(2b+1)+a(2c+1)}{3}\\ &= \frac{a+b+c+2(ab+bc+ca)}{3}\\ &\le \frac{(a+b+c)+\frac{2}{3}(a+b+c)^{2}}{3}. \end{aligned}\]In particular, if $(a+b+c=3)$, then
\[ba^{2/3}+cb^{2/3}+ac^{2/3} \le \frac{3+\frac{2}{3}\cdot 9}{3}=3,\]which proves (1), and we are done.
Equality holds when $(a=b=c=1)$. Therefore, 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
\[t_a<\frac{b+c}{2},\qquad t_b<\frac{a+c}{2},\qquad t_c<\frac{a+b}{2}.\]After adding these we get
\[t_a+t_b+t_c<a+b+c. \qquad (1)\]By squaring (1) we deduce
\[t_a^{2}+t_b^{2}+t_c^{2}+2(t_at_b+t_bt_c+t_ct_a) < a^{2}+b^{2}+c^{2}+2(ab+bc+ca). \qquad (2)\]On the other hand, we have
\[t_a^{2}=\frac{2(b^{2}+c^{2})-a^{2}}{4},\qquad t_b^{2}=\frac{2(a^{2}+c^{2})-b^{2}}{4},\qquad t_c^{2}=\frac{2(a^{2}+b^{2})-c^{2}}{4}.\]So
\[t_a^{2}+t_b^{2}+t_c^{2}=\frac{3}{4}(a^{2}+b^{2}+c^{2}).\]Using this and (2) we get
\[t_at_b+t_bt_c+t_ct_a < \frac{1}{8}(a^{2}+b^{2}+c^{2})+(ab+bc+ca). \qquad (3)\]Also, we have $(a^{2}+b^{2}+c^{2}<2(ab+bc+ca))$, since
\[a^{2}+b^{2}+c^{2}-2(ab+bc+ca) = a(a-b-c)+b(b-a-c)+c(c-a-b)<0.\]Finally, by (3) and the previous inequality we obtain
\[t_at_b+t_bt_c+t_ct_a<\frac{5}{4}(ab+bc+ca).\]Equality is never achieved for any non-degenerate triangle, since all inequalities used are strict. Therefore, the smallest constant 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
\[\begin{aligned} (a^{2}+b^{2}+c^{2})^{2} &=a^{4}+b^{4}+c^{4}+2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}\\ &=a^{4}+b^{4}+c^{4}+a^{2}(b^{2}+c^{2})+b^{2}(c^{2}+a^{2})+c^{2}(a^{2}+b^{2}). \qquad (1) \end{aligned}\]By Exercise 1.7, it follows that
\[a^{4}+b^{4}+c^{4}\ge abc(a+b+c). \qquad (2)\]Also,
\[b^{2}+c^{2}\ge 2bc,\qquad c^{2}+a^{2}\ge 2ca,\qquad a^{2}+b^{2}\ge 2ab. \qquad (3)\]Now by (1), (2), and (3) we deduce
\[\begin{aligned} (a^{2}+b^{2}+c^{2})^{2} &\ge abc(a+b+c)+2a^{2}bc+2b^{2}ac+2c^{2}ab\\ &=abc(a+b+c)+2abc(a+b+c)=3abc(a+b+c). \qquad (4) \end{aligned}\]Since (a+b+c\ge abc), from (4) we have
\[(a^{2}+b^{2}+c^{2})^{2}\ge 3abc(a+b+c)\ge 3(abc)^{2},\]i.e.
\[a^{2}+b^{2}+c^{2}\ge \sqrt{3}\,abc.\]Equality 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 $(a+b+c\ge abc)$ is satisfied with equality, and
\[a^{2}+b^{2}+c^{2}=3\cdot 3=9=\sqrt{3}\cdot 3\sqrt{3}=9.\]Therefore, 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
\[\begin{aligned} a^{4}+b^{4}+c^{4} &\ge abc(a+b+c)\\ &\Longleftrightarrow\ a^{4}+b^{4}+c^{4}\ge a^{2}bc+b^{2}ac+c^{2}ab\\ &\Longleftrightarrow\ \frac{T[4,0,0]}{2}\ge \frac{T[2,1,1]}{2}. \end{aligned}\]i.e.
\[T[4,0,0]\ge T[2,1,1],\]which is true according to Muirhead’s theorem.
Equality holds when $(a=b=c)$, since then both sides are equal:
\[a^{4}+b^{4}+c^{4}=3a^{4},\qquad abc(a+b+c)=a^{3}\cdot 3a=3a^{4}.\]Therefore, the largest constant is $(C=1)$.
Input: 2025.12.08 15:51