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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:

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

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

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    $\begingroup$ An explanation is needed for the first inequality. $\endgroup$
    – DeepSea
    Jun 28, 2016 at 8:34
  • $\begingroup$ @DeepSea: I think $$\sqrt[3]{\frac{a^3+b^3}2}\le \frac{a^2+b^2}{a+b}$$ $\endgroup$
    – Roman83
    Jun 28, 2016 at 9:02
  • $\begingroup$ But $2a \geq a+b$ is not true for all $a, b$. $\endgroup$
    – DeepSea
    Jun 28, 2016 at 9:05
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    $\begingroup$ But $\sum_{cyc}2a=\sum_{cyc}a+b$ for all $a,b,c$. $\endgroup$
    – Redundant Aunt
    Jun 28, 2016 at 9:14

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