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Question

Cho

$a;b;c>0$ .CMR:

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b}{b+c}+\frac{b+c}{a+b}+1$

Confusion · Answer 1 · 4/16/2016 6:29:40 PM

Ta có biến đổi:

$VT=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{a^2}{ab}+\frac{b^2}{bc}+\frac{c^2}{ca}\geq \frac{(a+b+c)^2}{ab+bc+ca}$ ( Bunhia.. phân thức)

$\rightarrow$ Ta chỉ cần chứng minh

$\frac{(a+b+c)^2}{ab+bc+ca} \ge \frac{a+b}{b+c}+\frac{b+c}{a+b}+1$

$\Leftrightarrow \frac{(a+b+c)^2}{ab+bc+ca}-3 \ge \frac{a+b}{b+c}+\frac{b+c}{a+b}-2$

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

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

$\Leftrightarrow (b^2-ac)^2 \ge0$ (luôn đúng)

$\Rightarrow$ đpcm , dấu

$"="$ xảy ra

$\Leftrightarrow a=b=c$

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