Cho các số thực dương $a$, $b$, $c$ thỏa mãn $a+b+c\geqslant ab+bc+ca.$ Chứng minh rằng $$\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\leqslant \dfrac{3}{2}.$$
$a$, $b$, $c>0$, $a+b+c\geqslant ab+bc+ca.$ Chứng minh $\sum \frac{ab}{a+b} \leqslant \frac{3}{2}.$
Started By HuyCubing, 17-05-2023 - 09:46
Best Answer huytran08, 09-06-2023 - 16:03
Ta sẽ chứng minh:$ \sum \frac{ab}{a+b}\leq \frac{3}{2}.\frac{ab+bc+ac}{a+b+c}$
Thật vậy:$\sum \frac{ab}{a+b}\leq \frac{3}{2}.\frac{ab+bc+ac}{a+b+c}\Leftrightarrow (a+b+c)\left ( \sum \frac{ab}{a+b} \right )\leq \frac{3}{2}(ab+bc+ac)$
$\Leftrightarrow \sum \frac{1}{a+b}\leq \frac{1}{2}\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )$ (luôn đúng)
$ \Rightarrow \sum \frac{ab}{a+b}\leq \frac{3}{2}.\frac{ab+bc+ac}{a+b+c}\leq \frac{3}{2}$
Dấu "=" xảy ra $\Leftrightarrow a=b=c=1$ (thỏa mãn)
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Posted 09-06-2023 - 16:03
huytran08
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Ta sẽ chứng minh:$ \sum \frac{ab}{a+b}\leq \frac{3}{2}.\frac{ab+bc+ac}{a+b+c}$
Thật vậy:$\sum \frac{ab}{a+b}\leq \frac{3}{2}.\frac{ab+bc+ac}{a+b+c}\Leftrightarrow (a+b+c)\left ( \sum \frac{ab}{a+b} \right )\leq \frac{3}{2}(ab+bc+ac)$
$\Leftrightarrow \sum \frac{1}{a+b}\leq \frac{1}{2}\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )$ (luôn đúng)
$ \Rightarrow \sum \frac{ab}{a+b}\leq \frac{3}{2}.\frac{ab+bc+ac}{a+b+c}\leq \frac{3}{2}$
Dấu "=" xảy ra $\Leftrightarrow a=b=c=1$ (thỏa mãn)
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