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## Inequality in triangle

We denote by $D,E,F$ the points where the incircle of triangle $ABC$ touches the sides $BC,CA,AB$, respectively. Prove that $\displaystyle \frac{BC}{EF}+\frac{AC}{FD}+\frac{AB}{DE}\geq 6$.
Solution: Denote by $p=(a+b+c)/2,\ x=p-a,\ y=p-b,\ z=p-c$. Then $EF=\frac{2xr}{\sqrt{x^2+r^2}}$, where $r$ is the radius if the incircle. We know that $r=\frac{S}{p}$, where $S$ is the area of the given triangle. Then, by Heron’s formula we get $r=\sqrt{\frac{xyz}{x+y+z}}$.
Replacing $EF, DE, DF$ we get that our inequality is equivalent to the following:
$\displaystyle \left(\sum_{cyc} \sqrt{\frac{x+y}{z}}\right)\sqrt{\frac{(x+y)(y+z)(z+x)}{xyz}}\geq 12$.
Use the AM-GM inequality to prove that the second part is greater than $2\sqrt{2}$.
For the other part, prove that $\displaystyle\sqrt{\frac{x+y}{z} \cdot 2} \geq 4 \frac{c}{a+b}$ and the other permutations, using the GM-HM inequality.
Remark: For $x_1,...,x_n >0$ we have:
$\displaystyle \frac{x_1+...+x_n}{n}\geq \sqrt[n]{x_1...x_n} \geq \frac{n}{\frac{1}{x_1}+...+\frac{1}{x_n}}$.
The first one is called AM-GM inequality, and the second one $GM-HM$ inequality.

$\displaystyle \frac{BC}{EF}+\frac{AC}{FD}+\frac{AB}{DE} \geq 3\sqrt[3]{\frac{AB\cdot BC\cdot CA}{EF\cdot FD\cdot ED}}=3 \sqrt[3]{\frac{RS}{rs}} \geq 6$
Which is equivalent to $RS \geq 8rs$. This is true, since it is kown that $R\geq 2r$ and $S\geq 4s$.
I guess that you denote by $s$ the area of the triangle $DEF$. Thank you for this very nice and geometrical solution.