Đáp án: $Q\ge\dfrac92$
Giải thích các bước giải:
Ta có:
$Q=x^2+\dfrac1x+y^2+\dfrac1y$
$\to Q=(x^2+y^2)+(\dfrac1x+\dfrac1y)$
$\to Q\ge \dfrac12(x+y)^2+\dfrac4{x+y}$
$\to Q\ge (\dfrac12(x+y)^2+\dfrac1{2(x+y)}+\dfrac1{2(x+y)})+\dfrac3{x+y}$
$\to Q\ge 3\sqrt[3]{\dfrac12(x+y)^2\cdot\dfrac1{2(x+y)}\cdot\dfrac1{2(x+y)}}+\dfrac3{1}$
$\to Q\ge \dfrac92$
Dấu = xảy ra khi $x+y=1$
