# An Inequality in Triangle with Radicals, Semiperimeter and Inradius

### Source

### Solution

By the Cauchy-Schwarz inequality,

$\displaystyle \begin{align} (LHS)^2 &= \left(\sum_{cycl}\sqrt{\frac{a+b}{(s-a)(s-b)}}\cdot\sqrt{s-a}\right)^2\\ &\le\frac{(a+b)(s-c)+(b+c)(s-a)+(c+a)(s-b)}{(s-a)(s-b)(s-c)}\cdot s. \end{align}$

But $\displaystyle \frac{s}{(s-a)(s-b)(s-c)}=\frac{1}{r^2}$ and

$\displaystyle \sum_{cycl}(a+b)(s-c)=a^2+b^2+c^2.$

by direct manipulation.

### Acknowledgment

The problem by Nguyen Viet Hung was shared at the CutTheKnotMath facebook page by Leo Giugiuc, along with a solution of his. Originally, the problem was posted at the Olimpiada pe Scoala (The School Yard Olympiad) facebook group.

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