# Michael Rozenberg's Inequality in Two Variables

### Solution 1

Due to the constraint $x+y=2,\,$ $xy\in [0,1].\,$ Let $xy=p.\,$ Then

$\displaystyle\sqrt{x^2+3}+\sqrt{y^2+3}=\sqrt{10-2p+2\sqrt{(p-3)^2+12}}$

so that the required inequality becomes

$\displaystyle\sqrt{10-2p+2\sqrt{(p-3)^2+12}}\ge 6-\sqrt{p+3}.$

Squaring gives

$\displaystyle 2\left(\sqrt{(p-3)^2+12}+6\sqrt{p+3}\right)-3p\ge 29.$

Let $f(p)\,$ denote the left-hand side. We have

$\displaystyle f'(p)=2\left(\frac{p-3}{\sqrt{(p-3)^2+12}}+\frac{3}{\sqrt{p+3}}\right)-3$

and

\displaystyle\begin{align} f''(p)&=6\left(\frac{4}{[(p-3)^2+12]^{3/2}}-\frac{1}{2(p+3)^{3/2}}\right)\\ &=6\left(\frac{\displaystyle 1}{\displaystyle\left[\frac{(p-3)^2+12}{4^{2/3}}\right]^{3/2}}-\frac{1}{[2^{2/3}(p+3)]^{3/2}}\right) \end{align}

We shall prove that $f''(0)\le 0,\,$ for $p\in [0,1].\,$ To this end, suffice it to show that, for $p\in [0,1],$

$\displaystyle \left[\frac{(p-3)^2+12}{4^{2/3}}\right]^{3/2}\ge [2^{2/3}(p+3)]^{3/2},$

or equivalently

$\displaystyle (p-3)^2+12-4(p+3)\ge 0,$

which is $(p-1)(p-9)\ge 0\,$ and obviously holds for $p\in [0,1].\,$ Thus it follows that $\max_{p\in [0,1]}f'(p)=f'(0)\lt 0,\,$ implying $\min_{p\in [0,1]}f(p)=f(1)=29,\,$ which completes the proof.

### Solution 2

Define, for $x\in[0,2],$

$f(x)=\sqrt{x^2+3}+\sqrt{x^2-4x+7}+\sqrt{-x^2+2x+3}.$

Then

$\displaystyle f'(x)=\frac{x}{\sqrt{x^2+3}}+\frac{x-2}{\sqrt{x^2-4x+7}}+\frac{1-x}{\sqrt{-x^2+2x+3}}.$

$f'=0\,$ for $x=1.\,$ Hence, $2\sqrt{3}+\sqrt{7}\ge f\ge6.$

By convexity, $x=y:$

### Acknowledgment

Leo Giugiuc has kindly communicated to me the above problem by Michael Rozenberg (Israel), originally posted at the artofproblemsolving forum, along with his solution.

The illustration is by Gary Davis.