# Inequalities with Constraint XV and XVI

### Problem 1, Solution 1

WLOG, we may assume that $a\ge b\ge c.\,$ It follows that $a^2\ge b^2\ge c^2\,$ and $\displaystyle\frac{1}{\sqrt{a^2+4}}\le\frac{1}{\sqrt{b^2+4}}\le\frac{1}{\sqrt{c^2+4}}.\,$ Now the Rearrangement inequality gives us

$\displaystyle\sum_{cycl}\frac{a^2}{\sqrt{b^2+4}}\ge\sum_{cycl}\frac{a^2}{\sqrt{a^2+4}}.$

It can be seen graphically (and proved with some tedious algebra/analysis) that the curve given by $\displaystyle y=\frac{x^2}{\sqrt{x^2+4}}\,$ lies on or above the tangent line to the curve at $x=1,\,$ $\displaystyle y=\frac{9\sqrt{5}}{25}(x-1)+\frac{\sqrt{5}}{5},\,$ on the interval $(0,3).\,$ Thus we have

\displaystyle\begin{align} \sum_{cycl}\frac{a^2}{\sqrt{b^2+4}} &\ge \sum_{cycl}\left(\frac{9\sqrt{5}}{25}(a-1)+\frac{\sqrt{5}}{5}\right)\\ &=\frac{9\sqrt{5}}{25}\sum_{cycl}-\frac{12\sqrt{5}}{25}\\ &=\frac{27\sqrt{5}}{25}-\frac{12\sqrt{5}}{25}=\frac{3\sqrt{5}}{5}\gt\frac{3}{5}. \end{align}

### Problem 1, Solution 2

Using Bergström's inequality and the obvious $x+2\gt\sqrt{x^2+4},$

\displaystyle\begin{align} \sum_{cycl}\frac{a^2}{\sqrt{b^2+4}} &\ge \frac{(a+b+c)^2}{\displaystyle\sum_{cycl}\sqrt{a^2+4}}\\ &\gt\frac{(a+b+c)^2}{\displaystyle\sum_{cycl}(a+2)}=\frac{9}{9}=1. \end{align}

### Problem 2, Solution

As above, using Bergström's inequality, the obvious $x^2+2\gt\sqrt{x^4+4}\,$ and $(x+y+z)^2\ge x^2+y^2+z^2,$

\displaystyle\begin{align} \sum_{cycl}\frac{a^2}{\sqrt{b^4+4}} &\ge \frac{(a+b+c)^2}{\displaystyle\sum_{cycl}\sqrt{a^4+4}}\\ &\gt\frac{9}{\displaystyle\sum_{cycl}(a^2+2)}\ge\frac{9}{(a+b+c)^2+6}\\ &=\frac{9}{15}=\frac{3}{5}. \end{align}

### Acknowledgment

The problem (Problem 1) with a solution (Problem 1, Solution 1) has been posted to the Romanian Mathematical Magazine by Henry Ricardo. A somewhat more complicated (but similar) problem has been posted by Dan Sitaru, with credits to Anish Ray.