$\frac{(a+b)^2}{ab}+$ $\frac{(c+b)^2}{cb}+$$\frac{(a+c)^2}{ac}$ $\geq9+2($ $\frac{a}{b+c}+ $ $\frac{b}
{a+c}+$ $\frac{c}{b+a})$
$\frac{(a+b)^2}{ab}+$ $\frac{(c+b)^2}{cb}+$$\frac{(a+c)^2}{ac}$ $\geq9+2($ $\frac{a}{b+c}+ $ $\frac{b}
{a+c}+$ $\frac{c}{b+a})$
Đáp án:
Giải thích các bước giải:
Áp dụng BĐT Cô si cho 3 số > 0 $: (x + y + z)(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}) ≥ 9$
Ta có :
$ [(a + b) + (b + c) + (c + a)].(\dfrac{1}{a + b} + \dfrac{1}{b + c} + \dfrac{1}{c + a}) ≥ 9$
$ ⇔ 2(a + b + c)(\dfrac{1}{a + b} + \dfrac{1}{b + c} + \dfrac{1}{c + a}) ≥ 9$
$ ⇔ \dfrac{a + b + c}{a + b} + \dfrac{a + b + c}{b + c} + \dfrac{a + b + c}{c + a} ≥ \dfrac{9}{2} $
$ ⇔ \dfrac{c}{a + b} + \dfrac{a}{b + c} + \dfrac{b}{c + a} ≥ \dfrac{3}{2} (*)$
Mặt khác áp dụng Cô si cho 2 số $x, y> 0$ :
$ (x + y)(\dfrac{1}{x} + \dfrac{1}{y}) ≥ 4 ⇔ \dfrac{1}{x} + \dfrac{1}{y} ≥ \dfrac{4}{x + y}$
Ta có :
$ c(\dfrac{1}{a} + \dfrac{1}{b}) + a(\dfrac{1}{b} + \dfrac{1}{c}) + b(\dfrac{1}{c} + \dfrac{1}{a})≥ \dfrac{4c}{a + b} + \dfrac{4a}{b + c} + \dfrac{4b}{c + a}$
$ ⇔ (\dfrac{a}{b} + \dfrac{b}{a}) + (\dfrac{b}{c} + \dfrac{c}{b}) + (\dfrac{c}{a} + \dfrac{a}{c}) ≥ 3 + \dfrac{2a}{b + c} + \dfrac{2b}{c + a} + \dfrac{2c}{a + b}$ ( theo $(*))$
$ ⇔ (\dfrac{a}{b} + \dfrac{b}{a} + 2) + (\dfrac{b}{c} + \dfrac{c}{b} + 2) + (\dfrac{c}{a} + \dfrac{a}{c} + 2) ≥ 9 + 2(\dfrac{a}{b + c} + \dfrac{b}{c + a} + \dfrac{c}{a + b})$
$ ⇔ \dfrac{(a + b)²}{ab} + \dfrac{(b + c)²}{bc} + \dfrac{(c + a)²}{ca} ≥ 9 + 2(\dfrac{a}{b + c} + \dfrac{b}{c + a} + \dfrac{c}{a + b}) (đpcm)$
Dấu $’=’$ xảy ra khi $a = b = c$