Giải thích các bước giải:

Ta có:

$P=\dfrac1{a^2+b^2}+\dfrac1{ab}+4ab$

$\to P=(\dfrac1{a^2+b^2}+\dfrac1{2ab})+\dfrac1{4ab}+(\dfrac1{4ab}+4ab)$

$\to P\ge\dfrac4{a^2+b^2+2ab}+\dfrac1{(a+b)^2}+2\sqrt{\dfrac1{4ab}\cdot 4ab}$

$\to P\ge\dfrac4{(a+b)^2}+\dfrac1{(a+b)^2}+2$

$\to P\ge\dfrac4{1^2}+\dfrac1{1^2}+2$

$\to P\ge 7$

Dấu = xảy ra khi $a=b=\dfrac12$