Question

cho a,b,c>0 thỏa mãn \(a^2+2b^2\le3c^2\)

CM: \(\frac{1}{a}+\frac{2}{b}\ge\frac{3}{c}\)

Answer 1

\(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\overset{BĐT\text{ }Cô-si}{\ge}\frac{9}{a+b+b}=\frac{9}{a+2b}\)

Áp dụng bất đẳng thức Bu-nhi-a-cốp-xki ta có:\(\left(a+2b\right)^2\le\left(1^2+\sqrt{2}^2\right)\left[a^2+\left(\sqrt{2}b\right)^2\right]=3\left(a^2+2b^2\right)\le9c^2\\ \Rightarrow a+2b\le3c\\ \Rightarrow\frac{1}{a}+\frac{2}{b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}a=b\\a^2+2b^2=3c^2\\\frac{a}{1}=\frac{\sqrt{2}b}{\sqrt{2}}\end{matrix}\right.\Rightarrow a=b=c\)