# Questions tagged [a.m.-g.m.-inequality]

For questions about proving and manipulating the AM-GM inequality. To be used necessarily with the [inequality] tag.

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### generalized f-mean/power mean inequalities

I'd like to prove the following inequality (which seems to be true by numerics) $$(p-1)\frac{a^2+b^2}{2} \leq \Big(\frac{a^p+b^p}{2}\Big)^{2/p}$$ for all $a,b\in [0,1]$ and $p\in [1,2]$. I'd ...
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### Given 4 numbers $a, b, c, d> 0,$ show $16\max\limits_{\bigcirc}\left \{ a^{3}+ 3bcd \right \}\!\geq\!\left ( a+ b+ c+ d \right )^{3}$

Given four positive numbers $a, b, c, d.$ Prove that $$16\max\left \{ a^{3}+ 3bcd, b^{3}+ 3cda, c^{3}+ 3dab, d^{3}+ 3abc \right \}\geq\left ( a+ b+ c+ d \right )^{3}$$ the way I think is using the ...
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### Minimize $z=e^{-x_{1}}+e^{-2 x_{2}}$ Subject to $\quad x_{1}+x_{2} \leq 1 \ , x_{1}, x_{2} \geq 0$ without KKT

I want to minimize $z=e^{-x_{1}}+e^{-2 x_{2}}$ subject to $\quad x_{1}+x_{2} \leq 1 \ , x_{1}, x_{2} \geq 0$ without using the KKT conditions. The restrain is on regular inequalities like Cauchy or ...
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### Question regarding a proof of a inequality

Given $a,b,c, \in (0,\infty)$, then the following inequality holds $$\sqrt{5a^2+12ab+7b^2}+\sqrt{5b^2+12bc+7c^2}+\sqrt{5c^2+12ca+7a^2} \leq 2 \sqrt6 (a+b+c)$$ What I've tried: First, I noticed that we ...
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### $\sum\sqrt{\frac{2a}{b+c}}\le\sqrt[3]{9\sum\frac{a}{b}}$

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}}\le\sqrt[3]{9\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$$ It is from ...
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### Better way to prove $\frac{1}{4a^2+3b^2+2c^2}\leq |\sqrt[3]{4}a+\sqrt[3]{2}b+c|$

Question:- Let $a,b$ and $c$ be integers, not all equal to $0$. Show that $$\frac{1}{4a^2+3b^2+2c^2}\leq |\sqrt[3]{4}a+\sqrt[3]{2}b+c|$$ This problem was proposed in a canadian journal. The presented ...
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### Prove inequality with $a;b;c \in R$
Let $a;b;c \in \mathbb{R}$ such that $a+b+c=0$ Prove that: $P=\dfrac{a-1}{a^2+8}+\dfrac{b-1}{b^2+8}+\dfrac{c-1}{c^2+8} \geq -\dfrac{3}{8}$ I tried to do this: \$\dfrac{8a-8}{a^2+8}+2+\dfrac{8b-8}{b^2+...