Question

cho 2 số dương x,y thỏa mãn \(\dfrac{1}{x}+\dfrac{1}{y}=2\)

tìm max\(Q=\dfrac{1}{x^4+y^2+2xy^2}+\dfrac{1}{y^4+x^2+2yx^2}\)

Answer 1

Ta có:

\(2=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\)

\(\Leftrightarrow xy\ge1\)

Theo đề bài thì

\(\dfrac{1}{x^4+y^2+2xy^2}+\dfrac{1}{y^4+x^2+2yx^2}\le\dfrac{1}{4\sqrt[4]{x^6y^6}}+\dfrac{1}{4\sqrt[4]{x^6y^6}}\le\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)

Answer 2