Đáp án:
Giải thích các bước giải:
Áp dụng BĐT Bunhiacopxki:
$(a+b+c)\left(\dfrac{a}{(b+c)^2}+\dfrac{b}{(c+a)^2}+\dfrac{c}{(a+b)^2} \right) \geq \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} \right)^2$ (1)
Lại có:
$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} =\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ac+bc} $
$⇒\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} \geq \dfrac{(a+b+c)^2}{2(ab+bc+ca)} \geq \dfrac{3(ab+bc+ca)}{2(ab+bc+ca)}$
$⇒\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$
$⇒\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} \right)^2 \geq \dfrac{9}{4}$ (2)
(1);(2) suy ra:
$(a+b+c)\left(\dfrac{a}{(b+c)^2}+\dfrac{b}{(c+a)^2}+\dfrac{c}{(a+b)^2} \right) \geq \dfrac{9}{4}$
$⇒\dfrac{a}{(b+c)^2}+\dfrac{b}{(c+a)^2}+\dfrac{c}{(a+b)^2} \geq \dfrac{9}{4(a+b+c)}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c$
