# Dan Sitaru's Inequality: From Three Variables to Many in Two Ways

### Problem

### Solution 1

$\displaystyle \begin{align} &(a+b)^3+2(a+b)^2+(a+b)\underbrace{\geq}_{AM-GM}4\sqrt[4]{(a+b)^3\cdot (a+b)^4\cdot (a+b)}\\ &\qquad\qquad=4(a+b)^2\underbrace{\geq}_{AM-GM} 4(2\sqrt{ab})^2=16ab\\ &(a+b)[(a+b)^2+2(a+b)+1]\geq 16ab\\ &\frac{16ab}{a+b}\leq (a+b)^2+2(a+b)+1\\ &\frac{ab}{a+b}\leq \Bigr(\frac{a+b+1}{4}\Bigr)^2\\ &\sqrt{\frac{ab}{a+b}}\leq \frac{a+b+1}{4}\\ &\sum_{cycl}\sqrt{\frac{ab}{a+b}}\leq \frac{1}{4}\sum (a+b+1)\\ &\frac{3}{2}\leq \frac{1}{4}\Biggl(\sum_{cycl} a+\sum_{cycl} a+3\Biggl)\\ &6\leq 2\sum_{cycl}a+3\\ &2\sum_{cycl}a\geq 3\\ &a+b+c\geq \frac{3}{2}. \end{align}$

### Solution 2

$\displaystyle \sqrt{\frac{(a+b)+(b+c)+(c+a)}{3}} \geq \frac{\sqrt{a+b}}{3} + \frac{\sqrt{b+c}}{3} + \frac{\sqrt{c+a}}{3}.$

By the AM-GM inequality,

$\displaystyle \sqrt{a+b}\geq2\sqrt{\frac{ab}{a+b}}~\text{and cyclic variants}$

Combining the two

$\displaystyle \begin{align}&\sqrt{\frac{2(a+b+c)}{3}} \geq \frac{2}{3}\left[\sqrt{\frac{ab}{a+b}}+\sqrt{\frac{bc}{b+c}}+\sqrt{\frac{ca}{c+a}}\right]=1\\ &\Rightarrow a+b+c \geq \frac{3}{2}. \end{align}$

### Solution 3

Since, by the AM-HM inequality,

$\displaystyle \sqrt{\frac{a+b}{2}}\ge\sqrt{\frac{ab}{a+b}}$

then $\displaystyle \sum_{cycl}\sqrt{a+b}\ge 3.\,$ But $\displaystyle 3\sqrt{\frac{2(a+b+c)}{3}}\ge\sum_{cycl}\sqrt{a+b}\ge 3,\,$ implying $\displaystyle a+b+c\ge\frac{3}{2}.\,$ Equality at $\displaystyle a=b=c=\frac{1}{2},\,$ which is obvious rom the proof.

### Solution 4

We have

$\displaystyle \sqrt{\frac{a b}{a+b}}\leq \frac{\frac{1}{2}(a+b)}{ \sqrt{a+b}}=\sqrt{\frac{a+b}{4}}.$

so

$\displaystyle \sqrt{\frac{a+b}{4}}+\sqrt{\frac{b+c}{4}}+\sqrt{\frac{a+c}{4}}\geq \sqrt{\frac{a b}{a+b}}+\sqrt{\frac{b c}{b+c}}+\sqrt{\frac{a c}{a+c}}=\frac{3}{2}.$

Since, by the power mean inequality, norm 1 $\geq$ norm $\frac{1}{2}$,

$\displaystyle \frac{1}{3} \left(\frac{a+b}{4}+\frac{a+c}{4}+\frac{b+c}{4}\right)\geq \left(\frac{1}{3} \left(\sqrt{\frac{a+b}{4}}+\sqrt{\frac{a+c}{4}}+\sqrt{\frac{b+c}{4}}\right)\right)^2$

we rewrite the inequality

$\displaystyle \sqrt{3} \sqrt{\frac{a+b}{4}+\frac{a+c}{4}+\frac{b+c}{4}}\geq \frac{\sqrt{a+b}}{2}+\frac{\sqrt{a+c}}{2}+\frac{\sqrt{b+c}}{2}\geq \frac{3}{2}$

and, finally, $\displaystyle a+b+c \geq \frac{3}{2}.$

### Extra

The original proof above admits an organic extension to any number of variables. E.g.,

Prove that, for $a_k \gt 0,\,$ $k=1,\ldots,n,\,$ that satisfy $\displaystyle \sum_{cycl}\sqrt{\frac{a_ka_{k+1}}{a_{k}+a_{k+1}}}=\frac{n}{2},\,$ where $a_{n+1}=a_1,\,$ the following inequality holds

$\displaystyle \sum_{cycl}a_k\ge\frac{n}{2}.$

It is also natural to extend the inequality in a symmetric way:

Prove that, for $a_k \gt 0,\,$ $k=1,\ldots,n,\,$ that satisfy $\displaystyle \sum_{j,k=1}^n\sqrt{\frac{a_ka_j}{a_{k}+a_{j}}}=\frac{n^2}{2},\,$ the following inequality holds

$\displaystyle \sum_{k=1}^na_k\ge\frac{n}{2}.$

### Acknowledgment

The problem, along with the solution, was kindly mailed to me by Dan Sitaru on a LaTeX file. The problem has been originally proposed at the Romanian Mathematical Magazine. Solution 2 is by Amit Itagi; Solution 3 is by Leo Giugiuc; Solution 4 is by N. N. Taleb.

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