Question

cho a,b,c ≥0.CMR

a+b+c ≥\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

Answer 1

áp dụng cô si ta có : \(\left\{{}\begin{matrix}a+b\ge2\sqrt{ab}\\b+c\ge2\sqrt{bc}\\c+a\ge2\sqrt{ca}\end{matrix}\right.\)

cộng quế theo quế ta có : \(2a+2b+2c\ge2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\)

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

Answer 2

Cách khác :3

\(a+b+c\text{≥}\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

⇔ \(2\left(a+b+c\right)\text{≥}2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\)

⇔ \(a-2\sqrt{ab}+b+b-2\sqrt{bc}+c+c-2\sqrt{ac}+a\text{ ≥}0\)

⇔\(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{a}-\sqrt{c}\right)^2\text{≥}0\left(luôn-đg\right)\)

\("="\text{⇔}a=b=c\)