Give: a;b;c>0 and \(a^2+b^2+c^2=1\). Prove that: \(\dfrac{1}{3-ab}+\dfrac{1}{3-bc}+\dfrac{1}{3-ca}\le\dfrac{3}{2}\)
Give: a;b;c>0 and \(a^2+b^2+c^2=1\). Prove that:
\(\dfrac{1}{3-ab}+\dfrac{1}{3-bc}+\dfrac{1}{3-ca}\le\dfrac{3}{2}\)
