# Small Change Makes Big Difference

### Solution

Observe that the function $\displaystyle \frac{1}{\sqrt{x}}\,$ is convex on $(0,],\,$ meaning $\displaystyle \frac{1}{2}(f(x)+f(y))\ge f\left(\frac{x+y}{2}\right).\,$ For $\displaystyle x=1+a^2-\frac{(a-b)^2}{2}\,$ and $\displaystyle y=1+b^2-\frac{(a-b)^2}{2}\,$ $\displaystyle x+y=2+a^2+b^2-(a^2+b^2-2ab)=2(1+ab).$

### Acknowledgment

The problem and the above solution have been shared by Marian Dinca on facebook.

Most likely the problem has been inspired by an earlier one:

$\displaystyle\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}\le\frac{2}{\sqrt{1+ab}}$

where without the extra terms $\displaystyle \frac{(a-b)^2}{2}\,$ the inverse inequality was true.

### Refinement

Leo Giugiuc has observed and proved a refinement of the above:

For $a,b\ge 0,$

$\displaystyle \left[1+a^2-\frac{(a-b)^2}{2}\right]\left[1+b^2-\frac{(a-b)^2}{2}\right]\le (1+ab)^2.$

For a proof, se $a+b=2\sqrt{s}\,$ and $ab=p,\,$ such that $s\ge p\ge 0.\,$ Further, $\displaystyle \frac{(a-b)^2}{2}=2(s-p).\,$ The inequality reduces to $[1+a^2-2(s-p)][1+b^2-2(s-p)]\le (1+p)^2.\,$ This is equivalent to

$4(s-p)^2+4(s-p)\le 8(1+2s-p)(s-p).$

Since $s-p\ge 0,\,$ suffice it to show that $s-p+1\le 2(1+2s-p),\,$ or $p\le s+1,\,$ which is obviously true. Equality is attained for $s=p,\,$ i.e., $a=b.$

This is indeed an improvement because

$\displaystyle \frac{1}{\displaystyle 1+a^2-\frac{(a-b)^2}{2}}+\frac{1}{\displaystyle 1+b^2-\frac{(a-b)^2}{2}}\ge \frac{2}{\displaystyle \sqrt[4]{\left[1+a^2-\frac{(a-b)^2}{2}\right]\left[1+b^2-\frac{(a-b)^2}{2}\right]}}$