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Question

cho a,b,c > 0. CMR: $\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}$ $\le\frac{a+b+c}{2}$

Diệu Huyền · Accepted Answer

Ta có: $\left(x+y\right)^2\ge4xy$

$\Rightarrow\frac{xy}{x+y}\le\frac{1}{4}\left(x+y\right)$

$\Rightarrow\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{1}{4}\left(a+b\right)+\frac{1}{4}\left(b+c\right)+\frac{1}{4}\left(c+a\right)$

$=\frac{a+b+c}{2}$

Dấu $"="$ xảy ra $\Leftrightarrow a=b=c$Câu trả lời: Ta có: $\left(x+y\right)^2\ge4xy$

$\Rightarrow\frac{xy}{x+y}\le\frac{1}{4}\left(x+y\right)$

$\Rightarrow\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{1}{4}\left(a+b\right)+\frac{1}{4}\left(b+c\right)+\frac{1}{4}\left(c+a\right)$

$=\frac{a+b+c}{2}$

Dấu $"="$ xảy ra $\Leftrightarrow a=b=c$

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