# An Inequality with Just Two Variables

### Solution 1

The required inequality is equivalent to

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

or,

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

Focusing on the left-hand side:

\displaystyle\begin{align} \left(\frac{a+b}{2ab}\right)\left(\sqrt{\frac{a^2+b^2}{2}}\right)&+\left(\frac{2ab}{a+b}\right)\left(\sqrt{\frac{2}{a^2+b^2}}\right)\\ &= \frac{1}{\sqrt{2(a^2+b^2)}}\left(\frac{a+b}{2ab}(a^2+b^2)+\frac{4ab}{a+b}\right)\\ &\le \frac{1}{\sqrt{2(a^2+b^2)}}\left(\frac{a+b}{2ab}(a^2+b^2)+(a+b)\right)\\ &\le \frac{1}{a+b}\left(\frac{a+b}{2ab}(a^2+b^2)+(a+b)\right)\\ &= \frac{a^2+b^2}{2ab}+1\\ &\le\frac{1}{2}\left(\frac{a}{b}+\frac{b}{a}\right)+\frac{1}{2}\left(\frac{a}{b}+\frac{b}{a}\right)\\ &=\frac{a}{b}+\frac{b}{a}, \end{align}

where we applied the AM-GM inequality.

### Solution 2

We know that

$\text{Harmonic mean}\;\le\;\text{Arithmetic mean}\;\le\;\text{Quadratic mean},$

implying

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

such that

(1)

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

Also,

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

such that

(2)

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

Multiplying (1) and (2) we get

\displaystyle\begin{align} &\left(\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\right)\left(\frac{a+b}{2ab} + \sqrt{\frac{2}{a^2+b^2}}\right)\\ &\qquad\qquad\le 2\sqrt{\frac{a^2+b^2}{2}}\left(\sqrt{\frac{a^2+b^2}{2}}\left(\frac{1}{ab}\right)+\sqrt{\frac{2}{a^2+b^2}}\right)\\ &\qquad\qquad=\frac{a^2+b^2}{ab}+2\\ &\qquad\qquad=\frac{(a+b)^2}{ab}. \end{align}

### Solution 3

WLOG, assume $a\le b.\;$ As before,

$\displaystyle 0\lt a\le\frac{2ab}{a+b}=\frac{2}{\displaystyle\frac{1}{a}+\frac{1}{b}}\le\frac{a+b}{2}\le\sqrt{\frac{a^2+b^2}{2}}\le b.$

We'll use Schweitzer's inequality:

$\displaystyle\left(\sum_{k=1}^{n}x_k\right)\left(\sum_{k=1}^{n}\frac{1}{x_k}\right)\le\frac{(m+M)^2n^2}{4mM},$

where $x_1,\ldots,x_n\in [m,M],\;$ $m\gt 0.\;$

with $n=2,\;$ $\displaystyle x_1=\frac{2ab}{a+b},\;$ $\displaystyle x_2=\sqrt{\frac{a^2+b^2}{2}},\;$ we directly get

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

### Acknowledgment

The problem above has been kindly posted to the CutTheKnotMath facebook page by Dan Sitaru, along with several solutions. Solution 1 is by Kevin Soto Palacios; Solution 2 by Soumava Chakraborty; Solution 3 is by Dan Sitaru. The illustration is by Nassim N. Taleb.

Inequalities in Two Variables