Question

cho \(a^2+2b^2\le3c^2\)

Chứng minh rằng \(\frac{1}{a}+\frac{2}{b}\ge\frac{3}{c}\)

Answer 1

Áp dụng bất đẳng thức cauchy- schawarz

\(\left(a^2+2b^2\right)3\ge\left(a+2b\right)^2\)

\(\Rightarrow a^2+2b^2\ge\frac{\left(a+2b\right)^2}{3}\)\(\Rightarrow\frac{\left(a+2b\right)^2}{3}\le3c^2\Leftrightarrow\left(a+2b\right)^2\le9c^2\Leftrightarrow a+2b\le3c\)

áp dụng bất đẳng thức Cauchy - schawarz dạng engel

\(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{4}{2b}\ge\frac{\left(1+2\right)^2}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\)