Prove inequality $\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\frac{c^3+d^3}{2}}+\sqrt[3]{\frac{d^3+a^3}{2}} \le 2(a+b+c+d)-4$

Question

Let $a,b,c,d$ positive real numbers, such that $$\frac1a+\frac1b+\frac1c+\frac1d=4.$$ Prove inequality $$\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\frac{c^3+d^3}{2}}+\sqrt[3]{\frac{d^3+a^3}{2}} \le 2(a+b+c+d)-4$$

My work so far:

$$1=\frac{4}{\frac1a+\frac1b+\frac1c+\frac1d}\le \sqrt[4]{abcd}\le\frac{a+b+c+d}{4}$$

score 0 · Accepted Answer · answered Jun 28 '16 at 08:22

0

$\sum\limits_{cyc}\left(2a-\sqrt[3]{\frac{a^3+b^3}{2}}\right)\geq\sum\limits_{cyc}\left(a+b-\frac{a^2+b^2}{a+b}\right)=2\sum\limits_{cyc}\frac{ab}{a+b}=2\sum\limits_{cyc}\frac{1}{\frac{1}{a}+\frac{1}{b}}\geq\frac{32}{\sum\limits_{cyc}\left(\frac{1}{a}+\frac{1}{b}\right)}=4$

answered Jun 28 '16 at 08:22

Michael Rozenberg

194,933

2

An explanation is needed for the first inequality. – DeepSea Jun 28 '16 at 08:34
@DeepSea: I think $$\sqrt[3]{\frac{a^3+b^3}2}\le \frac{a^2+b^2}{a+b}$$ – Roman83 Jun 28 '16 at 09:02
But $2a \geq a+b$ is not true for all $a, b$. – DeepSea Jun 28 '16 at 09:05
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But $\sum_{cyc}2a=\sum_{cyc}a+b$ for all $a,b,c$. – Redundant Aunt Jun 28 '16 at 09:14

Prove inequality $\sqrt[3]{\frac{a^3+b^3}{2}}+\sqrt[3]{\frac{b^3+c^3}{2}}+\sqrt[3]{\frac{c^3+d^3}{2}}+\sqrt[3]{\frac{d^3+a^3}{2}} \le 2(a+b+c+d)-4$

My work so far:

1 Answers1