Question

Cho a,b,c>0 thỏa mãn điều kiện : \(a^2+b^2+c^2=3\) C/m \(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge3\)

Answer 1

Xét \(P=\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)^2\)

\(P=\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}+2\left(a^2+b^2+c^2\right)=\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}+6\)

Áp dụng BĐT Cauchy, ta có:

\(\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}\ge2\sqrt{b^4}=2b^2\)

Tương tự, ta có: \(P=\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}+6\ge a^2+b^2+c^2+6=9\)

\(\Rightarrow P\ge3\)