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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.

  1. Michael
    January 30, 2012 at 3:03 pm

    There is more geometrical solution. By Cauchy 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.

    • January 30, 2012 at 3:26 pm

      I guess that you denote by s the area of the triangle DEF. Thank you for this very nice and geometrical solution.

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