Áp dụng BĐT Cauchy-Schwarz ta có:
\(A=\frac{a^4}{a\left(b+2c\right)}+\frac{b^4}{b\left(c+2a\right)}+\frac{c^4}{c\left(a+2b\right)}\)
\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{a\left(b+2c\right)+b\left(c+2a\right)+c\left(a+2b\right)}\)
\(=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{3\left(ab+bc+ca\right)}\)
\(\ge\frac{\left(ab+bc+ca\right)\left(\frac{\left(a+b+c\right)^2}{3}\right)}{3\left(ab+bc+ca\right)}\)
\(\ge\frac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\) (ĐPCM)
Xảy ra khi \(a=b=c=1\)