Question

Cho a,b,c>0.

(a^2+b^2+c^2)(1/a+b+1/b+c+1/c+a)>=3/2(a+b+c)

Answer 1

Đặt \(A=\left(a^2+b^2+c^2\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel, ta có:

\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{\left(1+1+1\right)^2}{a+b+b+c+c+a}\)\(=\frac{9}{2\left(a+b+c\right)}\)(1)

Áp dụng BĐT Cauchy-Schwarz, ta có:

\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)(2)

Nhân (1) và (2) vế theo vế\(\Rightarrow A\ge\frac{3\left(a+b+c\right)}{2}\)