Question

Cho \(x>y\ge0\). CMR: \(P=2017\left[\frac{x^4+y^4}{x^4-y^4}-\frac{xy}{x^2-y^2}+\frac{x+y}{2\left(x-y\right)}\right]\ge\frac{6051}{2}\)

Answer 1

Dat \(A=\frac{x^4+y^4}{x^4-y^4}-\frac{xy}{x^2-y^2}+\frac{x+y}{2\left(x-y\right)}\)

\(=\frac{2x^4+2y^4-2xy\left(x^2+y^2\right)+\left(x+y\right)^2\left(x^2+y^2\right)}{2x^4-2y^4}\)

\(=\frac{2x^4+2y^4+\left(x^2+y^2\right)\left[\left(x+y\right)^2-2xy\right]}{2x^4-2y^4}\)

\(=\frac{2x^4+2y^4+\left(x^2+y^2\right)^2}{2x^4-2y^4}\)

\(\Rightarrow A\ge\frac{2x^4+x^4}{2x^4}=\frac{3}{2}\)

\(\Rightarrow P=2017A\ge2017.\frac{3}{2}=\frac{6051}{2}\)

Dau '=' xay ra khi \(y=0\)