BĐT khó

T

thinhrost1

BDT được viết lại:

$\sqrt{\dfrac{2a}{a+b}}+ \sqrt{\dfrac{2b}{b+c}}+ \sqrt{\dfrac{2c}{c+a}} \le 3$

$\sqrt{\dfrac{2a}{a+b}}+ \sqrt{\dfrac{2b}{b+c}}+ \sqrt{\dfrac{2c}{c+a}} \le \sqrt{(a+c+b+a+c+b) [ \dfrac{2a}{(a+b)(a+c)} +\dfrac{2b}{(b+a)(b+c)}+\dfrac{2c}{(c+a)(c+b)}}$

Mà: $ \dfrac{a}{(a+b)(a+c)}+\dfrac{b}{(b+c)(b+a)}+\dfrac{c}{(b+c)(a+c)}\le \dfrac{9}{4(a+b+c)}\Leftrightarrow a(b-c)^{2}+b(c-a)^{2}+c(a-b)^{2}\ge 0(DPCM)$
 
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