Giải thích các bước giải:
Bài 1:
Ta có:
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{a+b}$
$=\dfrac{ab}{a}+\dfrac{ab}{b}+\dfrac{2}{a+b}$
$=b+a+\dfrac{2}{a+b}$
$=(\dfrac{a +b}{2}+\dfrac{2}{a+b})+\dfrac{a+b}{2}$
$\ge 2\sqrt{\dfrac{a +b}{2}\cdot\dfrac{2}{a+b}}+\dfrac{2\sqrt{ab}}{2}$
$=3$
Dấu = xảy ra khi $a=b=1$
Bài 2:
Ta có:
$\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge 2\sqrt{\dfrac{a^2}{b+c}\cdot \dfrac{b+c}{4}}=a$
$\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge 2\sqrt{\dfrac{b^2}{c+a}\cdot \dfrac{c+a}{4}}=b$
$\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge 2\sqrt{\dfrac{c^2}{a+b}\cdot \dfrac{a+b}{4}}=c$
$\to \dfrac{a^2}{b+c}+\dfrac{b+c}{4}+\dfrac{b^2}{c+a}+\dfrac{c+a}{4}+\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge a+b+c$
$\to \dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge \dfrac{a+b+c}{2}$
Dấu = xảy ra khi $a=b=c$
