Giải thích các bước giải:
Đặt $t=\dfrac{a}{b}+\dfrac{b}{a}$
$\to t=\dfrac{a^2+b^2}{ab}$
$\to |t|=|\dfrac{a^2+b^2}{ab}|$
$\to |t|=\dfrac{a^2+b^2}{|ab|}$
$\to |t|\ge\dfrac{2\sqrt{a^2\cdot b^2}}{|ab|}$
$\to |t|\ge\dfrac{2|ab|}{|ab|}$
$\to |t|\ge 2$
Ta có:
$\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+4-3(\dfrac{a}b+\dfrac ba)$
$=(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+2)+2-3(\dfrac{a}b+\dfrac ba)$
$=(\dfrac{a}b+\dfrac ba)^2+2-3(\dfrac{a}b+\dfrac ba)$
$=t^2+2-3t$
$=(t-1)(t-2)$
Mà $|t|\ge 2\to t\ge 2$ hoặc $t\le -2$
Nếu $t\ge 2\to (t-1)(t-2)\ge (2-1)(2-2)=0$
$\to \dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+4-3(\dfrac{a}b+\dfrac ba)\ge 0$
$\to \dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+4\ge 3(\dfrac{a}b+\dfrac ba)(1)$
Nếu $t\le -2$
$\to t-1\le -2-1<0$
$t-2\le -2-2<0$
$\to (t-1)(t-2)>0$
$\to \dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+4-3(\dfrac{a}b+\dfrac ba)> 0$
$\to \dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+4> 3(\dfrac{a}b+\dfrac ba)(2)$
Từ $(1), (2)\to \dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+4\ge 3(\dfrac{a}b+\dfrac ba)$
$\to đpcm$
