$\frac{a^{2}}{b^{2}}$ + $\frac{b^{2}}{a^{2}}$ + 4 $\geq$ 3( $\frac{a}{b}$ + $\frac{b}{a}$ ) Mình giải thế này được không ạ ? ($\frac{a}{b}$ + $\frac{

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 $(1), (2)\to \dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+4\ge 3(\dfrac{a}b+\dfrac ba)$

$\to đpcm$