3 replies to this topic

[minhhien2001]

$ab\geqslant 1.$C/m;$\frac{1}{a^2+1}+\frac{1}{b^2+1}\geqslant \frac{2}{ab+1}$

[rainbow99]

$ab\geqslant 1.$C/m;$\frac{1}{a^2+1}+\frac{1}{b^2+1}\geqslant \frac{2}{ab+1}$

http://diendantoanho...slant-frac21ab/

[Lee LOng]

Ta có:

BĐT$\frac{1}{a^{2}+1}-\frac{1}{ab+1}+\frac{1}{b^{2}+1}-\frac{1}{ab+1}\geq 0$

$\Leftrightarrow \frac{ab+1-a^{2}-1}{(a^{2}+1)(ab+1)}+\frac{ab+1-b^{2}-1}{(b^{2}+1)(ab+1)}\geq 0$

$\Leftrightarrow a(b-a)(b^{2}+1)+b(a-b)(a^{2}+1)\geq 0$

$\Leftrightarrow (a-b)^{2}(ab-1)\geq 0$

$\Rightarrow Đpcm$

[arsfanfc]

$\frac{1}{a^2+1}-\frac{1}{ab+1} \geq \frac{1}{ab+1}-\frac{1}{b^2+1} $

$<=> \frac{ab-a^2}{(1+a^2)(ab+1)} \geq {(b^2-ab)(ab+1)}$

$<=> \frac{(a-b)^2(ab-1)}{(a^2+1)(b^2+1)(ab+1)} \geq 0$ < luôn đúng >

Edited by arsfanfc, 02-04-2015 - 21:40.