Proof Writing & Problem SolvingEasy ShareProof Writing & Problem Solving · Easy problemA direct proof that a+b2≥ab\dfrac{a + b}{2} \ge \sqrt{ab}2a+b≥ab for a,b≥0a, b \ge 0a,b≥0 uses the identity a+b2−ab=(a−b)22\dfrac{a + b}{2} - \sqrt{ab} = \dfrac{(\sqrt a - \sqrt b)^2}{2}2a+b−ab=2(a−b)2. For a=4a = 4a=4 and b=9b = 9b=9, what is a+b2−ab\dfrac{a + b}{2} - \sqrt{ab}2a+b−ab?Check Show answer Walk through it step by step Practice more like this