Ta có \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{b+c+c+a+a+b}\)(Schwarz)
\(=\dfrac{a+b+c}{2}\)
\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)
Tương tự: \(\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge b\) ; \(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
Cộng vế:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Rightarrow...\)
Áp dụng BĐT Bunhiacopski, ta có:
\(\left[\left(\dfrac{a}{\sqrt{b+c}}\right)^2+\left(\dfrac{b}{\sqrt{c+a}}\right)^2+\left(\dfrac{c}{\sqrt{a+b}}\right)^2\right]\left[\left(\sqrt{b+c}\right)^2+\left(\sqrt{c+a}\right)^2+\left(\sqrt{a+b}\right)^2\right]\\ \ge\left(\dfrac{a}{\sqrt{b+c}}\cdot\sqrt{b+c}+\dfrac{b}{\sqrt{c+a}}\cdot\sqrt{c+a}+\dfrac{c}{\sqrt{a+b}}\cdot\sqrt{a+b}\right)^2\\ \Leftrightarrow\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)\cdot2\left(a+b+c\right)\ge\left(a+b+c\right)^2\\ \Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=2\)
Dấu "=" khi a=b=c
