Cho a,b,c > 0.
CMR: $\frac{a}{{a^2 + 1}} + \frac{b}{{b^2 + 1}} + \frac{c}{{c^2 + 1}} \le \frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}}$
$\sum {\frac{a}{{a^2 + 1}}} \le \sum {\frac{a}{{b + c}}} $
Bắt đầu bởi NLT, 12-04-2012 - 20:23
#1
Đã gửi 12-04-2012 - 20:23
GEOMETRY IS WONDERFUL !!!
Some people who are good at calculus think that they will become leading mathematicians. It's funny and stupid.
Nguyễn Lâm Thịnh
#2
Đã gửi 12-04-2012 - 20:40
$\frac{a}{a^{2}+1}\leq \frac{a}{2a}=\frac{1}{2}$
$\frac{a}{a^{2}+1}+\frac{b}{b^{2}+1}+\frac{c}{c^{2}+1}\leq \frac{3}{2}$
Mà theo Netbitt 3 biến thì
$\frac{3}{2}\leq \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}$
Cm hoàn tất
$\frac{a}{a^{2}+1}+\frac{b}{b^{2}+1}+\frac{c}{c^{2}+1}\leq \frac{3}{2}$
Mà theo Netbitt 3 biến thì
$\frac{3}{2}\leq \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}$
Cm hoàn tất
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