Question

Cho a,b,c > 0 . Biết \(a^2+b^2+c^2=1\)

Tìm GTNN của \(P=\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\)

Answer 1

\(\left(a+b+c\right)^2\ge0\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)\ge0\)

\(\Leftrightarrow1+2\left(ab+bc+ac\right)\ge0\)

\(\Leftrightarrow ab+bc+ac\ge\frac{1}{2}\)

\(\left(ab+bc+ac\right)^2\ge\frac{1}{4}\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2\left(abbc+bcac+abac\right)\ge\frac{1}{4}\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)\ge\frac{1}{4}\)

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