Question

Với a,b,c >0

1/a+1/b+1/c >= 1/(\(\sqrt{ab}\))+1/(\(\sqrt{bc}\))+1/(\(\sqrt{ac}\)

Answer 1

Với mọi a;b;c dương, ta luôn có:

\(\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2+\left(\frac{1}{\sqrt{b}}-\frac{1}{\sqrt{c}}\right)^2+\left(\frac{1}{\sqrt{c}}-\frac{1}{\sqrt{a}}\right)^2\ge0\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}-\frac{2}{\sqrt{ab}}+\frac{1}{b}+\frac{1}{c}-\frac{2}{\sqrt{bc}}+\frac{1}{c}+\frac{1}{a}-\frac{2}{\sqrt{ca}}\ge0\)

\(\Leftrightarrow\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\ge\frac{2}{\sqrt{ab}}+\frac{2}{\sqrt{bc}}+\frac{2}{\sqrt{ca}}\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\)

Dấu "=" xảy ra khi \(a=b=c\)