Bài này áp dụng BĐT Cauchy-Schwarz: \(\left(m^2+n^2+p^2\right)\left(x^2+y^2+z^2\right)\ge\left(mx+ny+pz\right)^2\)
Xét:
\(\left[\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2+\left(\sqrt{c}\right)^2\right].\left[\left(\frac{\sqrt{a}}{b+c}\right)^2+\left(\frac{\sqrt{b}}{c+a}\right)^2+\left(\frac{\sqrt{c}}{a+b}\right)^2\right]\ge\)
\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\)(1)
Xét: \(\left[\left(\sqrt{ab+ca}\right)^2+\left(\sqrt{bc+ab}\right)^2+\left(\sqrt{ca+bc}\right)^2\right].\left[\left(\frac{a}{\sqrt{ab+ca}}\right)^2+\left(\frac{b}{\sqrt{bc+ab}}\right)^2+\left(\frac{c}{\sqrt{ca+bc}}\right)^2\right]\ge\)
\(\left(a+b+c\right)^2\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)(2)
Xét \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{3}{2}\)(3)
Từ (1), (2), (3)
\(\Rightarrow\left(a+b+c\right)\left[\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\right]\ge\left(\frac{3}{2}\right)^2=\frac{9}{4}\)
\(\Rightarrow\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\ge\frac{9}{4\left(a+b+c\right)}\)
