if a>0,b>0,c>0 then prove that a/(b+c)+b/(a+c)+c/(a+b)>=3/2
Let $\displaystyle b+c=x, \ a+c=y, \ a+b=z$
Then $\displaystyle a=\dfrac{-x+y+z}{2}, \ b=\dfrac{x-y+z}{2}, \ c=\dfrac{x+y-z}{2}$
The inequality becomes
$\displaystyle \dfrac{-x+y+z}{2x}+\dfrac{x-y+z}{2y}+\dfrac{x+y-z}{2z}\geq \dfrac{3}{2}$
or $\displaystyle \dfrac{1}{2}\left[\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfra c{x}{z}+\dfrac{z}{x}\right)+\left(\dfrac{y}{z}+\df rac{z}{y}\right)-3\right]\geq\dfrac{3}{2}$
But it is well known that if $\displaystyle a>0$ then $\displaystyle a+\dfrac{1}{a}\geq 2$
Can you finish?