The power of substitution II
proving an inequality with two variables

Let $a=\tan x\,$ and $b=\tan y,\,$ with $\displaystyle 0\le x,y\le\frac{\pi}{4}.\,$ The required inequality becomes

$\displaystyle \cos x+\cos y\le 2\frac{\sqrt{\cos x\cos y}}{\sqrt{\cos (x-y)}},$

or, equivalently,

$\displaystyle (\cos x+\cos y)^2\cos (x-y)\le 4\cos x\cos y.$

Note that $\displaystyle \cos x+\cos y\le2\cos\left(\frac{x+y}{2}\right).\,$ Thus, suffice it to prove that

$\displaystyle \cos^2\left(\frac{x+y}{2}\right)\cos (x-y)\le\cos x\cos y.$

Recollect that $\displaystyle \cos^2\left(\frac{x+y}{2}\right)=\frac{1+\cos (x+y)}{2}\,$ and $\displaystyle \cos x\cos y=\frac{\cos (x-y)+\cos (x+y)}{2},\,$ so that we need to prove

$\cos (x-y)+\cos (x-y)\cos (x+y)\le \cos(x-y)+\cos (x+y),$

which is obvious since $\cos (x+y)\ge 0\,$ and $\cos (x-y)\le 1.$ That completes the proof. Equality holds if $x=y,\,$ i.e., if $a=b.$

Note that

$\displaystyle \frac{1}{1+a^2}+\frac{1}{1+b^2}-\frac{2}{1+ab}=\frac{(ab-1)(a-b)^2}{(a^2+1)(b^2+1)(1+ab)}\le 0,$

because $ab\le 1.\,$ Equality is attained for $a=b.$ It follows that

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

By the Cauchy-Schwarz inequality,

$\sqrt{(a+a^2)(1+b^2)}\ge 1+ab,$

so that

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

and, subsequently,

$\displaystyle \left(\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}\right)^2\le \left(\frac{2}{\sqrt{1+ab}}\right)^2,$

or, equivalently,

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

Equality is attained for $a=b=1\,$ or $a=b=0.$

Let $\displaystyle f(x)=\sqrt{1+e^{2x}},\,$ $x\le 0.\,$ $\displaystyle f'(x)=e^{2x}\left(1+e^{2x}\right)^{-\frac{3}{2}},$

$\displaystyle \begin{align} f''(x) &= -2e^{2x}\left(1+e^{2x}\right)^{-\frac{3}{2}}+3e^{4x}\left(1+e^{2x}\right)^{-\frac{5}{2}}\\ &=e^{2x}\left(1+e^{2x}\right)^{-\frac{5}{2}}\left[3e^{2x}-2(1+e^{2x})\right]\\ &=e^{2x}\left(1+e^{2x}\right)^{-\frac{5}{2}}(e^{2x}-2)\lt 0, \end{align}$

since $x\lt 0.\,$ Thus, $f(x)\,$ is concave and, according to Jensen's inequality, $\displaystyle f(x)+f(y)\le 2f\left(\frac{x+y}{2}\right),\,$ or

$\displaystyle (1+e^{2x})^{-\frac{1}{2}}+(1+e^{2y})^{-\frac{1}{2}}\le 2(1+e^{x+y})^{-\frac{1}{2}}.$

Substituting $e^x=a\,$ and $a^y=b\,$ gives $\displaystyle (1+a^2)^{-\frac{1}{2}}+(1+b^2)^{-\frac{1}{2}}\le 2(1+ab)^{-\frac{1}{2}},\,$ which is

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

Leo Giugiuc has kindly posted a problem from the artofproblemsolving forum at the CutTheKnotMath facebook page, along with a beautiful and instructive solution (Solution 1). No less instructive was the purely algebraic solution (Solution 2) by Imad Zak, while Marian Dinca has employed a different kind of substitution (Solution 3).