$\frac{1}{(c+a)^2(a+b)^2}+\frac{1}{(a+b)^2(b+c)^2}+\frac{1}{(b+c)^2(c+a)^2}\leq \frac{2}{(ab+bc+ca)^2}$
Started By no matter what, 15-12-2012 - 19:00
#1
Posted 15-12-2012 - 19:00
Chứng minh với mọi a,b,c dương
$\frac{1}{(c+a)^2(a+b)^2}+\frac{1}{(a+b)^2(b+c)^2}+\frac{1}{(b+c)^2(c+a)^2}\leq \frac{2}{(ab+bc+ca)^2}$
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#2
Posted 21-12-2012 - 23:03
:
Chứng minh với mọi a,b,c dương
$\frac{1}{(c+a)^2(a+b)^2}+\frac{1}{(a+b)^2(b+c)^2}+\frac{1}{(b+c)^2(c+a)^2}\leq \frac{2}{(ab+bc+ca)^2}$
Đặt $a+b+c=p;ab+bc+ca=q;abc=r$
BĐT $\frac{(a+b)^2+(b+c)^2+(c+a)^2}{(a+b)^2(b+c)^2(c+a)^2}\leq \frac{2}{(ab+bc+ca)^2}$
$\Leftrightarrow \frac{a^2+b^2+c^2+ab+bc+ca}{(a+b)^2(b+c)^2(c+a)^2}\leq \frac{1}{(ab+bc+ca)^2}$
$\Leftrightarrow \frac{p^2-q}{(pq-r)^2}\leq \frac{1}{q^2}$
$\Leftrightarrow q^2(p^2-q)\leq (pq-r)^2$
$\Leftrightarrow q^3+r^2\geq 2pqr$
(Cái này luon đúng do $q^2\geq 2pr$ )
Đẳng thức xảy ra chẳng hạn khi $b=c\rightarrow 0$ và $a\neq 0$
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