Question

[Baek Hyun]

Cho a,b,c >0 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)=2016. CMR: \(\frac{bc}{a^2\left(3b+c\right)}+\frac{ca}{b^2\left(3c+a\right)}+\frac{ab}{c^2\left(3a+b\right)}\ge504\)

Answer 1

[Trần Phúc Khang]

Ta có

\(VT=\frac{\frac{1}{a^2}}{\frac{3}{c}+\frac{1}{b}}+\frac{\frac{1}{b^2}}{\frac{3}{a}+\frac{1}{c}}+\frac{\frac{1}{c^2}}{\frac{3}{b}+\frac{1}{a}}\)

Áp dụng bất đẳng thức buniacoxki dạng phân thức:

=> \(VT\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{4}{a}+\frac{4}{b}+\frac{4}{c}}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{4}=504\)

Dấu bằng xảy ra khi a=b=c=3/2016