IMO LONGLISTED PROBLEM 1970

[MATHPRITOM]

$\frac{bc}{b+c} $+$\frac{ca}{c+a}$+$\frac{ab}{a+b}$ $\ge$ $\frac{a+b+c}{2}$.where a,b,c>0.

[MATHPRITOM]

a very nice problem.

[Masum]

Let, $a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}$
Then the inequality is reduced to \[\sum \frac{1}{x+y}\ge \frac{xy+yz+zx}{xyz}=\sum \frac{1}{2z}\]
This is true by re-arrangement inequality.

[MATHPRITOM]

thanks 4 a very nice,small solution.my solution is too long.