Let $a$, $b$, $c$, $k$ be positive reals such that $ab+bc+ca \leq 1$ and $0 < k \leq \frac{9}{2}$. Prove that: \[\sqrt[3]{ \frac{k}{a} + (9-3k)b} + \sqrt[3]{\frac{k}{b} + (9-3k)c} + \sqrt[3]{\frac{k}{c} + (9-3k)a } \leq \frac{1}{abc}.\] Proposed by Zhang Yanzong and Song Qing
Problem
Source: Olympic Revenge 2021 #1
Tags: Inequality, algebra, inequalities, olympic revenge
12.11.2021 12:46
jpedrorvc1 wrote: Let $a$, $b$, $c$, $k$ be positive reals such that $ab+bc+ca \leq 1$ and $0 < k \leq \frac{9}{2}$. Prove that: \[\sqrt[3]{ \frac{k}{a} + (9-3k)b} + \sqrt[3]{\frac{k}{b} + (9-3k)c} + \sqrt[3]{\frac{k}{c} + (9-3k)a } \leq \frac{1}{abc}.\] Proposed by Zhang Yanzong and Song Qing Thank jpedrorvc1. Let $a,b,c$ be three positive real numbers such that $ab+bc+ca = 1.$ Prove that $$\sqrt[3]{ \frac{2}{a} +3b} + \sqrt[3]{\frac{2}{b} +3c} + \sqrt[3]{\frac{2}{c} +3a } \leq \frac{1}{abc}$$$$\sqrt[3]{ \frac{4}{a}-3b} + \sqrt[3]{\frac{4}{b}-3c} + \sqrt[3]{\frac{4}{c}-3a } \leq \frac{1}{abc}$$h h
19.06.2022 12:21
jpedrorvc1 wrote: Let $a$, $b$, $c$, $k$ be positive reals such that $ab+bc+ca \leq 1$ and $0 < k \leq \frac{9}{2}$. Prove that: \[\sqrt[3]{ \frac{k}{a} + (9-3k)b} + \sqrt[3]{\frac{k}{b} + (9-3k)c} + \sqrt[3]{\frac{k}{c} + (9-3k)a } \leq \frac{1}{abc}.\] Proposed by Zhang Yanzong and Song Qing
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