# An Inequality in Triangle with Radicals, Semiperimeter, Incenter and Inradius

### Solution

Introduce $x=\sqrt{s-a},$ $y=\sqrt{s-b},$ $z=\sqrt{s-c}.$ Then $a=y^2+z^2,$ $b=z^2+x^2,$ $c=x^2+y^2.$ With $\displaystyle AI^2=\frac{(s-a)bc}{s},$ $\displaystyle BI^2=\frac{(s-b)ca}{s},$ $\displaystyle CI^2=\frac{(s-c)ab}{s},$ and $\displaystyle r^2=\frac{(s-a)(s-b)(s-c)}{s},$ the required inequality reduces to

$\displaystyle 3+\sum_{cycl}\frac{\sqrt{(x^2+y^2)(x^2+z^2)}}{yz}\ge\frac{(x+y+z)(xy+yz+zx)}{xyz}.$

The latter simplifies to

$\displaystyle 3xyz+\sum_{cycl}x\sqrt{(x^2+y^2)(x^2+z^2)}\ge (x+y+z)(xy+yz+zx).$

By the Cauchy-Schwarz inequality,

\displaystyle \begin{align} 3xyz+\sum_{cycl}x\sqrt{(x^2+y^2)(x^2+z^2)}&\ge 3xyz+\sum_{cycl}x(xy+zx)\\ &=(x+y+z)(xy+yz+zx). \end{align}

### Acknowledgment

The problem by Lorian Saceanu was kindly shared at the CutTheKnotMath facebook page by Leo Giugiuc, along with a solution of his. Originally, the problem was posted at the Easy Beautiful Math facebook group.

Copyright © 1996-2018 Alexander Bogomolny

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