Giải thích các bước giải:
3.Ta có:
$A=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+4-3(\dfrac{a}{b}+\dfrac{b}{a})$
$\to A=(\dfrac{a^2}{b^2}+2\cdot \dfrac ab\cdot \dfrac ba+\dfrac{b^2}{a^2})+2-3(\dfrac{a}{b}+\dfrac{b}{a})$
$\to A=(\dfrac{a}{b}+\dfrac{b}{a})^2+2-3(\dfrac{a}{b}+\dfrac{b}{a})$
$\to A=(\dfrac{a}{b}+\dfrac{b}{a})^2-3(\dfrac{a}{b}+\dfrac{b}{a})+2$
$\to A=(\dfrac{a}{b}+\dfrac{b}{a}-1)(\dfrac{a}{b}+\dfrac{b}{a}-2)$
Đặt $\dfrac ab+\dfrac ba =t $
$\to A=(t-1)(t-2)$
$\to |t|=|\dfrac ab+\dfrac ba|=|\dfrac{a^2+b^2}{ab}|=\dfrac{a^2+b^2}{|ab|}\ge \dfrac{2|ab|}{|ab|}=2$
$\to t\ge 2$ hoặc $t\le -2$
Nếu $t\ge 2$
$\to t-1\ge 1, t-2\ge 0\to A=(t-1)(t-2)\ge 0$
Nếu $t\le -2\to t-1\le -3, t-2\le -4\to (t-1)(t-2)>0$
Kết hợp cả $2$ trường hợp
$\to A\ge 0$
$\to \dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+4-3(\dfrac{a}{b}+\dfrac{b}{a})\ge0$
$\to \dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+4\ge 3(\dfrac{a}{b}+\dfrac{b}{a})$
5.Ta có:
$a^4+b^4\ge \dfrac12(a^2+b^2)^2$
$\to a^4+b^4\ge \dfrac12(\dfrac12(a+b)^2)^2$
$\to a^4+b^4\ge \dfrac18(a+b)^4$
$\to \dfrac{a^4+b^4}{2}\ge (\dfrac{a+b}{2})^4$
