Áp dụng BĐT Svac - xơ:
\(\frac{1}{a^2+2ab}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\)\(=\frac{1^2}{a^2+2ab}+\frac{1^2}{b^2+2ac}+\frac{1^2}{c^2+2ab}\)
\(\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}=\frac{9}{\left(a+b+c\right)^2}\ge9\)(Vì \(a+b+c\le1\))
(Dấu "="\(\Leftrightarrow a=b=c=\frac{1}{3}\))