Mình thấy phương pháp này đúng còn yếu khi cm một số bài mạnh, nhưng lên mathlinks thấy có 2 bài anh Hùng làm hay quá, post thêm cho nó phong phú.
Prove that for any nonnegative real numbers we have
$ \dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+\dfrac{9(ab+bc+ca)}{a^2+b^2+c^2} \geq 12 $
Rewrite the inequality to
$ (a-b)^{2}\left(\dfrac{(a+b)^{2}}{a^{2}b^{2}}-\dfrac{9}{a^{2}+b^{2}+c^{2}}\right)+(c-a)(c-b)\left(\dfrac{(c+a)(c+b)}{a^{2}c^{2}}-\dfrac{9}{a^{2}+b^{2}+c^{2}}\right). $
Let
$ M=\dfrac{(a+b)^{2}}{a^{2}b^{2}}-\dfrac{9}{a^{2}+b^{2}+c^{2}} $
$ N=\dfrac{(c+a)(c+b)}{a^{2}c^{2}}-\dfrac{9}{a^{2}+b^{2}+c^{2}}. $
We may assume that $ a\ge b\ge c $ , then $ N\ge 0 $ (it is very interesting). If $ a-b\le b-c $ then
$ (c-a)(c-b) \ge 2(a-b)^{2} $
and the inequality is proved if we can prove that
$ M+2N \ge \dfrac{(c+a)(c+b)}{a^{2}c^{2}}-\dfrac{10}{a^{2}+b^{2}+c^{2}}\ge 0. $
Indeed, this inequality is equivalent to
$ (c+a)(c+b)(a^{2}+b^{2}+c^{2}) \ge 10a^{2}c^{2}, $
Since $ b\ge \dfrac{a+c}{2} $, we refer
$ (c+a)(c+b)(a^{2}+b^{2}+c^{2}) \ge (c+a)\left(c+\dfrac{a+c}{2}\right)\left(a^{2}+c^{2}+\dfrac{1}{4}(a+c)^{2}\right) $
$ \ge 2\sqrt{ac}\cdot\sqrt{3ac}\cdot 3ac>10ac. $
So it remains to consider the case $ a-b\ge b-c $. In this case, we will prove that $M\ge 0$ or
$ (a+b)^{2}(a+b+c)^{2}\ge 9a^{2}b^{2}. $
If $ a\ge 2b $ then this inequality is true immediately because $ a^{2}+b^{2}\ge \dfrac{5}{2}ab $. So we may assume that $ a\le 2b $ . Because $ c\ge 2b-a $, we will prove next that
$ (a+b)^{2}\left(a^{2}+b^{2}+(2b-a)^{2}\right)^{2}\ge 9a^{2}b^{2}. $
Let $ x=\dfrac{a}{b}, then 1\le x\le 2 $. The inequality becomes
$ (x+1)^{2}(2x^{2}-4x+5) \ge 9x^{2} $
or
$ 2x^{4}-10x^{2}+6x+5 \ge 0. $
This can be check easily. We have done.
Remark. The inequality $ N\ge 0 $ can be proved due to (sinc $ c\le b $)
$ (c+a)(2c)(a^{2}+2c^{2}) \ge 9c^{2}a^{2} $
or
$ 2a^{3}-7a^{2}c+4c^{2}a+4c^{3}\ge 0 $
or
$ (a-c)^{2}(2a+c) \ge 0. $
Bài viết đã được chỉnh sửa nội dung bởi 10maths_tp0609: 08-05-2007 - 17:34
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