Cho x,y,z khác 1,xyz=1.
CM:$\sum \frac{x^2}{(x-1)^2}\geq 1$
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Cho x,y,z khác 1,xyz=1.
CM:$\sum \frac{x^2}{(x-1)^2}\geq 1$
Cách khác:
Đặt $x=\frac{a^2}{bc}, y=\frac{b^2}{ca}, z=\frac{c^2}{ab}$.
BĐT trở thành $\sum \frac{a^4}{(a^2-bc)^2} \geq 1$
Áp dụng Cauchy-Schwarz: $\sum \frac{a^4}{(a^2-bc)^2} \geq \frac{(a^2+b^2+c^2)^2}{\sum (a^2-bc)^2}$
Cần chứng minh $(a^2+b^2+c^2)^2 \geq \sum (a^2-bc)^2$
$$a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2) \geq a^4+b^4+c^4+a^2b^2+b^2c^2+c^2a^2-2abc(a+b+c)$$
$$(ab+bc+ca)^2 \geq 0$$
BĐT hiển nhiên đúng.
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