Question

Cho hai số thực dương x,y thỏa mãn \(x+y+1=3xy\)

Tìm GTLN của:

\(M=\dfrac{3x}{y\left(x+1\right)}+\dfrac{3y}{x\left(y+1\right)}-\dfrac{1}{x^2}-\dfrac{1}{y^2}\)

Answer 1

Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\Rightarrow a+b+ab=3\)

Ta có: \(3-a+b+ab\ge ab+2\sqrt{ab}\ge3.\sqrt[3]{a^2b^2}\Leftrightarrow ab\le1\)

Suy ra \(M=\dfrac{ab}{a+1}+\dfrac{ab}{b+1}=ab.\left(\dfrac{a+1+b+1}{ab+a+b+1}\right)=ab.\dfrac{5-ab}{4}\)

\(=\dfrac{-\left[\left(ab\right)^2-2ab+1\right]+3a+1}{4}=\dfrac{-\left(ab-1\right)^2+3ab+1}{4}\le1\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=1\)