# Dan Sitaru's Cyclic Inequality In Three Variables with Constraints

### Solution 1

Since $\displaystyle y=\frac{1}{\sqrt{x}}\,$ is a convex function,

\displaystyle \begin{align} \sum_{cycl}\frac{1}{\sqrt{a+b^2}} &\ge 3\left(\frac{\displaystyle \sum_{cycl}a+\sum_{cycl}a^2}{3}\right)^{-\frac{1}{2}}\\ &= 3\left(\frac{27(a+b+c)}{3}\right)^{-\frac{1}{2}}\\ &=(a+b+c)^{-\frac{1}{2}}. \end{align}

### Solution 2

(1)

$\displaystyle LHS\ge\frac{9}{\sqrt{a+b^2}+\sqrt{b+c^2}+\sqrt{c+a^2}}.$

Now, using $\displaystyle \left(\sum_{cycl}x\right)^2\le 3\sum_{cycl}x^2,$

\displaystyle \begin{align} \left(\sum_{cycl}\sqrt{a+b^2}\right)^2&\le 3\left(\sum_{cycl}a+\sum_{cycl}a^2\right)\\ &=81\left(\sum_{cycl}a\right). \end{align}

So to continue from (1),

\displaystyle\begin{align} LHS&\ge\frac{9}{\sqrt{a+b^2}+\sqrt{b+c^2}+\sqrt{c+a^2}}\\ &\ge\frac{9}{9\sqrt{a+b+c}}\\ &=\frac{1}{\sqrt{a+b+c}}. \end{align}

### Solution 3

By Hölder's inequality,

\displaystyle\begin{align} \left(\sum_{cycl}\frac{1}{\sqrt{a+b^2}}\right)&\left(\sum_{cycl}\frac{1}{\sqrt{a+b^2}}\right)\left(\sum_{cycl}(a+b^2)\right)\\ &\ge\left(\sum_{cycl}\frac{1}{\sqrt[6]{a+b^2}}\cdot\frac{1}{\sqrt[6]{a+b^2}}\cdot\sqrt[3]{a+b^2}\right)^3\\ &=27. \end{align}

In other words,

$\displaystyle \left(\sum_{cycl}\frac{1}{\sqrt{a+b^2}}\right)^2\left(\sum_{cycl}a+\sum_{cycl}a^2\right)\ge 27,$

implying,

\displaystyle\begin{align} \left(\sum_{cycl}\frac{1}{\sqrt{a+b^2}}\right)^2&\ge\frac{27}{\displaystyle \sum_{cycl}a+26\sum_{cycl}a}\\ &=\frac{27}{\displaystyle 27\sum_{cycl}a}=\frac{1}{\displaystyle\sum_{cycl}a}. \end{align}

This is equivalent to the required inequality.

### Solution 4

Applying the AM-HM inequality,

\displaystyle\begin{align} \sum_{cycl}\frac{1}{\sqrt{a+b^2}}&\ge \frac{9}{\sqrt{(1+1+1)(a+b+c+a^2+b^2+c^2)}}\\ &=\frac{9}{\sqrt{3(27(a+b+c))}}\\ &=\frac{9}{\sqrt{9\cdot 9(a+b+c)}}\\ &=\frac{1}{a+b+c}\\ \end{align}

### Acknowledgment

Dan Sitaru has kindly posted the CutTheKnotMath facebook page the above problem from his book "Algebraic Phenomenon". Solution 1 is by Maki Milanovic; Solution 2 is by; Solution 3 is by Dan Sitaru; Solution 4 is by Shivam Sharma.