Giải thích các bước giải:

Ta có:

$M=(x^2-\dfrac1{y^2})(y^2-\dfrac1{x^2})$ 

$\to M=\dfrac{(x^2y^2-1)^2}{x^2y^2}$

Đặt $x^2y^2=t$

Ta có: $xy\le\dfrac14(x+y)=\dfrac14$

$\to x^2y^2\le\dfrac1{16}$

$\to t\le\dfrac1{16}$

Ta có:

$M=\dfrac{(t-1)^2}t$

$\to M=\dfrac{t^2-2t+1}t$

$\to M=t-2+\dfrac1t$

$\to M=t+\dfrac1{16^2t}-2+\dfrac1t-\dfrac1{16^2t}$

$\to M=t+\dfrac1{16^2t}-2+\dfrac{255}{256t}$

$\to M\ge 2\sqrt{t\cdot \dfrac1{16^2t}}-2+\dfrac{255}{256\cdot \dfrac1{16}}$

$\to M\ge\dfrac{225}{16}$

Dấu = xảy ra khi $t=\dfrac{1}{16^2t}$ và $t=\dfrac1{16}$

$\to t=\dfrac1{16}$

$\to xy=\dfrac14$

$\to \begin{cases}x+y=1\\xy=\dfrac14\end{cases}$

$\to x=y=\dfrac12$