Anonymous

Given that a, b & c>0. Proof: (a-b)^2>0 => a^2+b^2>2ab => a^2+b^2+2ab>4ab => (a+b)^2>4ab => (a+b)/(ab)>4/(a+b) => 1/a+1/b>4/(a+b) => 1/a+1/b>2/(a+b)

Anonymous

(1/a) + (1/b) > 2/(a + b); b + a > 2ab/(a + b) (b + a)^2 > 2ab; (b^2) + 2ab + (a^2) > 2ab; (a^2) + (b^2) > 0; proof is complete,