$Cho$ $a,b,c>0$. $CM$ $\frac{a}{(b+c)²}$ + $\frac{b}{(a+c)²}$ + $\frac{c}{(a+b)²}$ ≥ $\frac{9}{4(a+b+c)}$

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1. 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 $a=b=c$

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