$$\frac{a}{b}
+ \frac{b}{c} + \frac{c}{a} \geqslant \frac{{a + 2012}}{{b + 2012}} +
\frac{{b + 2012}}{{c + 2012}} + \frac{{c + 2012}}{{a + 2012}}$$
Giả sử $c= min{a,b,c}$
Ta có: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=(\frac{a}{b}+\frac{b}{a}-2)+(\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-1)$$
$$=\frac{a^2+b^2-2ab}{ab}+\frac{ab+c^2-bc-ac}{ac}$$
$$=\frac{(a-b)^2}{ab}+\frac{(a-c)(b-c)}{ac}$$
Tương tự ta có: $$\frac{a+2012}{b+2012}+\frac{b+2012}{c+2012}+\frac{c+2012}{a+2012}-3$$
Giả sử $c= min{a,b,c}$
Ta có: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=(\frac{a}{b}+\frac{b}{a}-2)+(\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-1)$$
$$=\frac{a^2+b^2-2ab}{ab}+\frac{ab+c^2-bc-ac}{ac}$$
$$=\frac{(a-b)^2}{ab}+\frac{(a-c)(b-c)}{ac}$$
Tương tự ta có: $$\frac{a+2012}{b+2012}+\frac{b+2012}{c+2012}+\frac{c+2012}{a+2012}-3$$
$$=\frac{(a-b)^2}{(a+2012)(b+2012)}+\frac{(a-c)(b-c)}{(c+2012)(a+2012)}$$
vì $a\geq c,b\geq c$
$$\Rightarrow \frac{(a-b)^2}{ab}+\frac{(a-c)(b-c)}{ac}\geq \frac{(a-b)^2}{(a+2012)(b+2012)}+\frac{(a-c)(b-c)}{(c+2012)(a+2012)}$$
Do đó $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\geq \frac{a+2012}{b+2012}+\frac{b+2012}{c+2012}+\frac{c+2012}{a+2012}-3$$
Cộng 2 vế cho 3 ta có đpcm
vì $a\geq c,b\geq c$
$$\Rightarrow \frac{(a-b)^2}{ab}+\frac{(a-c)(b-c)}{ac}\geq \frac{(a-b)^2}{(a+2012)(b+2012)}+\frac{(a-c)(b-c)}{(c+2012)(a+2012)}$$
Do đó $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\geq \frac{a+2012}{b+2012}+\frac{b+2012}{c+2012}+\frac{c+2012}{a+2012}-3$$
Cộng 2 vế cho 3 ta có đpcm
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