# A Cyclic Inequality of Degree Four

### Problem

### Solution

Observe that the inequality is equivalent to

$\displaystyle \sum_{cycl}b(a^4-a\sqrt{3}+2)\ge 0.$

But

$\displaystyle\begin{align} a^4-a\sqrt{3}+2&=(a^2-1)^2+2a^2-a\sqrt{3}+1\\ &=(a^2-1)^2+\left(\sqrt{2}a-\sqrt{\frac{3}{8}}\right)^2+\frac{5}{8}\\ &\gt 0. \end{align}$

### Extension

Note that the original inequality is weak, for example,

$\displaystyle\begin{align} a^4-a\sqrt{5}+2&=(a^2-1)^2+2a^2-a\sqrt{5}+1\\ &=(a^2-1)^2+\left(\sqrt{2}a-\sqrt{\frac{5}{8}}\right)^2+\frac{3}{8}\\ &\gt 0 \end{align}$

Shows that $\displaystyle \sum_{cycl}a^4b+2\sum_{cycl}a\ge \sqrt{5}\sum_{cycl}ab\,$ is also true which organically leads to the question of finding the maximum $k\,$ such that

$\displaystyle a^4b+b^4c+c^4a+2(a+b+c)\ge k(ab+bc+ca).$

Simply following the derivation above, the maximal $k\,$ satisfies, $k\ge k_0=2\sqrt{2}.$ But we can do better.

Let $f(x)=x^4-xk+2.\,$ Then $f'(x)=4x^3-k=0\,$ leads to $\displaystyle x=\sqrt[3]{\frac{k}{4}}.\,$ It is easy to see that this is a local minimum because $f''\left(\sqrt[3]{\frac{k}{4}}\right)\ge 0.$

Now, $\displaystyle f\left(\sqrt[3]{\frac{k}{4}}\right)=\left(\sqrt[3]{\frac{k}{4}}\right)^4-k\sqrt[3]{\frac{k}{4}}+2=2-3\left(\sqrt[3]{\frac{k}{4}}\right)^4.\,$ From this, $\displaystyle k_1=4\left(\frac{2}{3}\right)^{\frac{3}{4}}.\,$ Now, this is an imporvement since $k_1/k_0\approx 1.0434.\,$

The graph (courtesy wolframalpha) shows that this may not be the last word:

However, the graph generated by GeoGebra tells us a different story:

### Acknowledgment

Dan Sitaru has kindly posted the above problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and Leo Giugiuc messaged his solution practically right away.

Jeffrey Samuelson gave an answer to the extended question.

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