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Root inequality

Let a, b, c, d be positive real numbers satisfying the following equality

    \[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} =4\]

Prove that

\displaystyle \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}} \leq 2 \left ( a+b+c+d \right ) -4

Solution

We begin by stating a lemma:

Lemma: Let a, b be positive real numbers, then:

    \[\frac{a+b}{2} \leq \sqrt[3]{\frac{a^3+b^3}{2}} \leq \frac{a^2+b^2}{a+b}\]

Now, making use of the lemma we have that:

    \begin{align*} \sum \sqrt[3]{\frac{a^3+b^3}{2}} &\leq \sum \frac{a^2+b^2}{a+b} \\ &=\sum \left ( a+b \right )-\sum \frac{2ab}{a+b} \\ &= 2\left ( a+b+c+d \right )- 2\sum \frac{1}{\frac{1}{a}+\frac{1}{b}} \end{align*}

Making use of the Cauchy – Schwartz inequality we have that

    \begin{align*} \sum \frac{1}{\frac{1}{a}+\frac{1}{b}} & \geq \frac{\left ( 1+1+1+1 \right )^2}{2\left ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}+ \frac{1}{d} \right )} \\ &=\frac{8}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}+ \frac{1}{d}} \\ &= 2 \end{align*}

The inequality now follows.

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