Question

Cho \(a,b,c>0;a+b+c\ge3\).CMR : 

\(\frac{a^3}{b+2c}+\frac{b^3}{c+2a}+\frac{c^3}{a+2b}\ge1\)

 

Answer 1

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 \(a=b=c=1\)