Inequality Proof Problems [351-400]
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Restructured the IneqMath training data.
P351. Let $a, b \in \mathbb{R}^{+}$. Find the smallest constant $C$ such that the following inequality holds for all $a, b \in \mathbb{R}^{+}$: \(C(a^3 + b^3) \geq (a + b)^3\)
S351. $C = 4$
The function $f(x)=x^{3}$ is convex on $(0,+\infty)$, thus from Jensen’s inequality it follows that\n\(\n\\left(\\frac{a+b}{2}\\right)^{3} \\leq \\frac{a^{3}+b^{3}}{2} \\Leftrightarrow 4\\left(a^{3}+b^{3}\\right) \\geq(a+b)^{3}\n\)\n\nEquality holds when $a = b$, since then $a^3 + b^3 = 2a^3$ and $(a + b)^3 = (2a)^3 = 8a^3$, so $4(a^3 + b^3) = 8a^3 = (a + b)^3$. This gives the minimum value of $C$, so $C = 4$ is the smallest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 4$.
P352. Let $x, y, z > 0$ be real numbers. Determine the maximal constant $C$ such that the following inequality holds for all $x, y, z$: \((x y + y z + z x) \left( \frac{1}{(x + y)^2} + \frac{1}{(y + z)^2} + \frac{1}{(z + x)^2} \right) \geq C.\)
S352. $C = \frac{9}{4}$
The given inequality is equivalent to\n\(\n\\begin{align*}\n& 4(x y+y z+z x)\\left((z+x)^{2}(y+z)^{2}+(x+y)^{2}(z+x)^{2}+(x+y)^{2}(y+z)^{2}\\right) \\\\\n& \\quad \\geq 9(x+y)^{2}(y+z)^{2}(z+x)^{2} \\tag{1}\n\\end{align*}\n\)\n\nLet us denote $p=x+y+z, q=x y+y z+z x, r=x y z$.\nBy $I_{5}$ and $I_{7}$ we have\n\(\n(x+y)^{2}(y+z)^{2}(z+x)^{2}=(p q-r)^{2}\n\)\nand\n\(\n(x+y)^{2}(y+z)^{2}+(y+z)^{2}(z+x)^{2}+(z+x)^{2}(x+y)^{2}=\\left(p^{2}+q\\right)^{2}-4 p(p q-r)\n\)\n\nSo we can rewrite inequality (1) as follows\n\(\n\\begin{aligned}\n& 4 q\\left(\\left(p^{2}+q\\right)^{2}-4 p(p q-r)\\right) \\geq 9(p q-r)^{2} \\\\\n& \\quad \\Leftrightarrow \\quad 4 p^{4} q-17 p^{2} q^{2}+4 q^{3}+34 p q r-9 r^{2} \\geq 0 \\\\\n& \\quad \\Leftrightarrow \\quad 3 p q\\left(p^{3}-4 p q+9 r\\right)+q\\left(p^{4}-5 p^{2} q+4 q^{2}+6 p r\\right)+r(p q-9 r) \\geq 0\n\\end{aligned}\n\)\n\nThe last inequality follows from $N_{1}, N_{2}$ and $N_{3}$, and the fact that $p, q, r>0$. Equality occurs if and only if $x=y=z$.\n\nEquality holds when $x = y = z$, in which case $(xy + yz + zx) = 3x^2$ and $\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2} = \frac{3}{(2x)^2} = \frac{3}{4x^2}$, so the left side is $3x^2 \cdot \frac{3}{4x^2} = \frac{9}{4}$. Thus, the minimum value of the expression is $C = \frac{9}{4}$.\n\nTherefore, the answer is $C = \frac{9}{4}$.
P353. Let $a, b, c$ be the lengths of the sides of a triangle. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$: \((a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c} \leq C (a^{a} b^{b} c^{c}).\)
S353. $C = 1$
By the weighted power mean inequality we have\n\(\n\\begin{aligned}\n& \\sqrt[a+b+c]{\\left(\\frac{a+b-c}{a}\\right)^{a}\\left(\\frac{b+c-a}{b}\\right)^{b}\\left(\\frac{c+a-b}{c}\\right)^{c}} \\\\\n& \\leq \\frac{1}{a+b+c}\\left(a \\cdot \\frac{a+b-c}{a}+b \\cdot \\frac{b+c-a}{b}+c \\cdot \\frac{c+a-b}{c}\\right)=1\n\\end{aligned}\n\)\ni.e.\n\(\n(a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c} \\leq a^{a} b^{b} c^{c}\n\)\n\nEquality occurs when $a = b = c$, that is, for an equilateral triangle. In this case, both sides of the inequality are equal, so $C = 1$ is the minimal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.
P354. Let $x, y, z \in \mathbb{R}^{+}$ such that $x y z = 1$ and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq x + y + z$. Determine the largest constant $C$ such that for any natural number $n$, the following inequality holds for all $x, y, z$: \(\frac{1}{x^{n}} + \frac{1}{y^{n}} + \frac{1}{z^{n}} \geq C (x^{n} + y^{n} + z^{n}).\)
S354. $C = 1$
After setting $x=\frac{a}{b}, y=\frac{b}{c}$ and $z=\frac{c}{a}$, the initial condition\n\(\n\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z} \\geq x+y+z\n\)\nbecomes\n\(\n\\begin{aligned}\n\\frac{b}{a} & +\\frac{c}{b}+\\frac{a}{c} \\geq \\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a} \\\\\n& \\Leftrightarrow \\quad a^{2} b+b^{2} c+c^{2} a \\geq a b^{2}+b c^{2}+c a^{2} \\\\\n& \\Leftrightarrow \\quad(a-b)(b-c)(c-a) \\leq 0\n\\end{aligned}\n\)\n\nLet $n \in \mathbb{N}$, and take $A=a^{n}, B=b^{n}, C=c^{n}$.\nThen $a \geq b \Leftrightarrow A \geq B$ and $a \leq b \Leftrightarrow A \leq B$, etc.\nSo we have\n\(\n\\begin{aligned}\n&(A-B)(B-C)(C-A) \\leq 0 \\\\\n& \\Leftrightarrow \\quad \\frac{B}{A}+\\frac{C}{B}+\\frac{A}{C} \\geq \\frac{A}{B}+\\frac{B}{C}+\\frac{C}{A} \\\\\n& \\Leftrightarrow \\quad \\frac{1}{x^{n}}+\\frac{1}{y^{n}}+\\frac{1}{z^{n}} \\geq x^{n}+y^{n}+z^{n}\n\\end{aligned}\n\)\nExpected Answer: $C = 1$\n\nEquality holds when $a = b = c$, which corresponds to $x = y = z = 1$. In this case, $\frac{1}{x^{n}} + \frac{1}{y^{n}} + \frac{1}{z^{n}} = 3$ and $x^{n} + y^{n} + z^{n} = 3$, so the inequality becomes $3 \geq C \cdot 3$, i.e., $C = 1$. This gives the minimum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.
P355. Let $x, y, z \in \mathbb{R}^{+}$ such that $x + y + z = 1$. Determine the minimal constant $C$ such that the following inequality holds for all $x, y, z$: \(\frac{1}{1-x y}+\frac{1}{1-y z}+\frac{1}{1-z x} \leq C.\)
S355. $C = \frac{27}{8}$
Let $p=x+y+z=1, q=x y+y z+z x, r=x y z$.\nIt can easily be shown that\n\(\n(1-x y)(1-y z)(1-z x)=1-q+p r-r^{2}\n\)\nand\n\(\n(1-x y)(1-y z)+(1-y z)(1-z x)+(1-z x)(1-x y)=3-2 q+p r\n\)\n\nSo the given inequality becomes\n\(\n\\begin{gathered}\n8(3-2 q+p r) \\leq 27\\left(1-q+p r-r^{2}\\right) \\\\\n\\Leftrightarrow \\quad 3-11 q+19 p r-27 r^{2} \\geq 0\n\\end{gathered}\n\)\n\nSince $p=1$, we need to show that\n\(\n3-11 q+19 r-27 r^{2} \\geq 0\n\)\n\nBy $N_{5}: p^{3} \geq 27 r$ we have $1 \geq 27 r$, i.e. $r \geq 27 r^{2}$.\n\nTherefore\n\(\n3-11 q+19 r-27 r^{2} \\geq 3-11 q+19 r-r=3-11 q+18 r\n\)\n\nSo it suffices to prove that\n\(\n3-11 q+18 r \\geq 0\n\)\n\nWe have\n\(\n\\begin{aligned}\n& 3-11 q+18 r \\geq 0 \\\\\n& \\quad \\Leftrightarrow \\quad 3-11(x y+y z+z x)+18 x y z \\geq 0 \\\\\n& \\quad \\Leftrightarrow \\quad 11(x y+y z+z x)-18 x y z \\leq 3\n\\end{aligned}\n\)\n\nApplying $A M \geq G M$ we deduce\n\(\n\\begin{aligned}\n11(x y+y z+z x)-18 x y z & =x y(11-18 z)+11 z(x+y) \\\\\n& \\leq \\frac{(x+y)^{2}}{4}(11-18 z)+11 z(x+y) \\\\\n& =(1-z)^{2}/4(11-18 z)+11 z(1-z) \\\\\n& =\\frac{(1-z)((1-z)(11-18 z)+44 z)}{4} \\\\\n& =\\frac{4 z+3 z^{2}-18 z^{3}+11}{4}\n\\end{aligned}\n\)\n\nSo it remains to show that\n\(\n\\begin{aligned}\n& \\frac{4 z+3 z^{2}-18 z^{3}+11}{4} \\leq 3 \\\\\n& \\quad \\Leftrightarrow \\quad 4 z+3 z^{2}-18 z^{3} \\leq 1 \\\\\n& \\Leftrightarrow \\quad 18 z^{3}-3 z^{2}-4 z+1 \\geq 0 \\\\\n& \\Leftrightarrow \\quad(3 z-1)^{2}(2 z+1) \\geq 0\n\\end{aligned}\n\)\nwhich is obvious.\n\nEquality holds when $x = y = z = \frac{1}{3}$, since then $x + y + z = 1$ and $x y = y z = z x = \frac{1}{9}$, so $\frac{1}{1-x y} = \frac{1}{1-1/9} = \frac{9}{8}$, and the sum is $3 \times \frac{9}{8} = \frac{27}{8}$. This gives the maximum value of the sum, so $C = \frac{27}{8}$ is the minimal constant such that the inequality always holds.\n\nTherefore, the answer is $C = \frac{27}{8}$.
P356. Let $\alpha_{i}>0, i=1,2, \ldots, n$, be real numbers such that $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=1$. Determine the maximal constant $C$ such that the following inequality holds for all $\alpha_{i}$: \(\alpha_{1}^{\alpha_{1}} \alpha_{2}^{\alpha_{2}} \cdots \alpha_{n}^{\alpha_{n}} \geq C.\)
S356. $C = \frac{1}{n}$
If we take $a_{i}=\frac{1}{\alpha_{i}}, i=1,2, \ldots, n$, by the Weighted $AM-GM$ inequality we get\n\(\n\\frac{1}{\\alpha_{1}^{\\alpha_{1}}} \\frac{1}{\\alpha_{2}^{\\alpha_{2}}} \\cdots \\frac{1}{\\alpha_{n}^{\\alpha_{n}}} \\leq \\frac{1}{\\alpha_{1}} \\alpha_{1}+\\frac{1}{\\alpha_{2}} \\alpha_{2}+\\cdots+\\frac{1}{\\alpha_{n}} \\alpha_{n}=n\n\)\ni.e.\n\(\n\\frac{1}{n} \\leq \\alpha_{1}^{\\alpha_{1}} \\alpha_{2}^{\\alpha_{2}} \\cdots \\alpha_{n}^{\\alpha_{n}}\n\)\nEquality holds when $\alpha_1 = \alpha_2 = \cdots = \alpha_n = \frac{1}{n}$, in which case $\alpha_1^{\alpha_1} \cdots \alpha_n^{\alpha_n} = \left(\frac{1}{n}\right)^1 = \frac{1}{n}$. This gives the minimum value of the product, so $C = \frac{1}{n}$ is maximal.\n\nTherefore, the answer is $C = \frac{1}{n}$.
P357. Let $a, b \in \mathbb{R}^{+}$ and $n \in \mathbb{N}$. Find the smallest constant $C$ such that the following inequality holds for all $a, b \in \mathbb{R}^{+}$ and $n \in \mathbb{N}$: \((a+b)^{n}\left(a^{n}+b^{n}\right) \leq C\left(a^{2 n}+b^{2 n}\right)\)
S357. $C = 2^n$
By the power mean inequality, for any $x, y \in \mathbb{R}^{+}, n \in \mathbb{N}$, we have\n\(\n\\left(\\frac{x+y}{2}\\right)^{n} \\leq \\frac{x^{n}+y^{n}}{2}\n\)\n\nTherefore\n\(\n\\begin{aligned}\n(a+b)^{n}\\left(a^{n}+b^{n}\\right) & =2^{n}\\left(\\frac{a+b}{2}\\right)^{n}\\left(a^{n}+b^{n}\\right) \\\\\n& \\leq 2^{n}\\left(\\frac{a^{n}+b^{n}}{2}\\right)\\left(a^{n}+b^{n}\\right)=2^{n} \\frac{\\left(a^{n}+b^{n}\\right)^{2}}{2} \\\\\n& \\leq 2^{n} \\frac{2\\left(a^{2 n}+b^{2 n}\\right)}{2}=2^{n}\\left(a^{2 n}+b^{2 n}\\right)\n\\end{aligned}\n\)\n\nEquality holds when $a = b$, since all the inequalities above become equalities in this case. This gives the minimum value of $C$, so $C = 2^n$ is the smallest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 2^n$.
P358. Let $a_{1}, a_{2}, \ldots, a_{n} \in \mathbb{R}^{+}$. Find the smallest constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$: \(\sum_{k=1}^{n} k a_{k} \leq C + \sum_{k=1}^{n} a_{k}^{k}\)
S358. $C = \binom{n}{2}$
For $1 \leq k \leq n$ we have\n\(\na_{k}^{k}+(k-1)=a_{k}^{k}+\\underbrace{1+1+\\cdots+1}_{k-1} \\geq k \\sqrt[k]{a_{k}^{k}}=k a_{k}\n\)\n\nAfter adding these inequalities, for $1 \leq k \leq n$ we get\n\(\n\\sum_{k=1}^{n} k a_{k} \\leq \\sum_{k=1}^{n} a_{k}^{k}+\\sum_{k=1}^{n}(k-1)=\\sum_{k=1}^{n} a_{k}^{k}+\\frac{n(n-1)}{2}=\\sum_{k=1}^{n} a_{k}^{k}+\\binom{n}{2}\n\)\n\nEquality holds when $a_k = 1$ for all $k$, since then $a_k^k = 1$ and $k a_k = k$, so both sides are equal. This gives the minimum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = \binom{n}{2}$.
P359. Let $a, b, c$ be positive numbers. Consider the following inequality: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \quad () \quad \frac{5 a}{2 a^2+b^2+c^2}+\frac{5 b}{a^2+2 b^2+c^2}+\frac{5 c}{a^2+b^2+2 c^2}.\)
Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S359. (F) None of the above
P360. Let $a, b, c$ be positive real numbers such that $4abc = a + b + c + 1$. Consider the following inequality: \(\frac{b^2+c^2}{2a}+\frac{c^2+a^2}{3b}+\frac{b^2+a^2}{c} \quad () \quad 2(ab+bc+ca).\)
Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S360. (F) None of the above
P361. Let $a, b, c > 0$ be positive real numbers such that $a+b+c \geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$. Consider the following inequality: \(\frac{a}{2a+1}+\frac{b}{b+1}+\frac{c}{c+1} \quad () \quad 2.\)
Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S361. (F) None of the above
P362. Let $a, b, c$ be positive real numbers such that $abc = 1$. Consider the following inequality: \(\sum \frac{1}{b(a+b)} \quad () \quad \frac{5}{3}.\)
Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S362. (F) None of the above
P363. Let $a, b, c$ be positive real numbers such that $abc = 1$. Consider the following inequality: \(\sum \frac{a(a^3+1)}{3b+c} \quad () \quad 2a^3 + b^3 + c^3.\)
Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S363. (F) None of the above
P364. Let $a, b, c$ be non-negative numbers such that $ab + ac + bc \neq 0$. Consider the following inequality: \(a^3 + b^3 + c^3 + 2abc + \frac{4a^2b^2c^2}{a^3 + b^3 + c^3 + abc} \quad () \quad \sum_{\text{cyc}}(2a^2b + 3a^2c) .\)
Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S364. (F) None of the above
P365. Let $a, b, c \in \mathbb{R}^{+}$. Consider the following inequality: \(\sqrt{\frac{a+b}{2c}} + \sqrt{\frac{b+c}{2a}} + \sqrt{\frac{c+a}{b}} \quad () \quad \sqrt{\frac{8(a+b+c)}{\sqrt[3]{abc}}}.\)
Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S365. (F) None of the above
P366. Let $a, b, c \geq 0$ be non-negative real numbers such that $a + b + c = 1$. Consider the following inequality: \(a^3 + b^3 + c^3 + 6abc \quad () \quad \frac{1}{3}.\)
Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S366. (F) None of the above
P367. Let $a, b, c \neq 0$ be non-zero real numbers such that $a^2 + b^2 + c^2 = 2(ab + bc + ca)$. Consider the following expression: \((a+b+c)\left(\frac{1}{2a}+\frac{1}{3b}+\frac{1}{c}\right) \quad () \quad 14.\)
Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S367. (F) None of the above
P368. Let $a, b, c \geq 0$ be non-negative real numbers such that $a^2 + b + c^2 = 1$. Consider the following inequality: \(\frac{1}{1+a^2} + \frac{1}{3(1+b)} + \frac{1}{1+c^2} \quad () \quad \frac{7}{4}.\)
Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S368. (F) None of the above
Input: 2025.12.08 15:51