Đáp án: $P\ge \dfrac92$
Giải thích các bước giải:
Ta có:
$2ab-4=a+b$
$\to 2(a+b)=4ab-8\le (a+b)^2-8$
$\to (a+b)^2-2(a+b)-8\ge 0$
$\to (a+b-4)(a+b+2)\ge 0$
$\to a+b-4\ge 0$ vì $a,b>0$
$\to a+b\ge 4$
$\to ab=\dfrac{a+b+4}{2}$
$\to P=\dfrac{a+b+4}{2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}$
$\to P=\dfrac{a+b}{2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+2$
$\to P\ge \dfrac{a+b}{2}+\dfrac12(\dfrac1a+\dfrac1b)^2+2$
$\to P\ge \dfrac{a+b}{2}+\dfrac12(\dfrac4{a+b})^2+2$
$\to P\ge \dfrac{a+b}{2}+\dfrac{8}{(a+b)^2}+2$
$\to P\ge \dfrac{a+b}{2}+(\dfrac{8}{(a+b)^2}+\dfrac12)++\dfrac32$
$\to P\ge \dfrac{a+b}{2}+2\sqrt{\dfrac{8}{(a+b)^2}\cdot\dfrac12}+\dfrac32$
$\to P\ge \dfrac{a+b}{2}+\dfrac{4}{a+b}+\dfrac32$
$\to P\ge \dfrac{a+b}{4}+(\dfrac{a+b}{4}+\dfrac{4}{a+b})+\dfrac32$
$\to P\ge \dfrac{4}{4}+2\sqrt{\dfrac{a+b}{4}\cdot\dfrac{4}{a+b}}+\dfrac32$
$\to P\ge \dfrac92$
Dấu = xảy ra khi $a=b=2$
