Cho $a,b,c$ là các số thực dương. Chứng minh $\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}\leq 3$
Chứng minh $\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}\leq 3$
Started By chanlerscofield, 12-07-2016 - 23:54
#1
Posted 12-07-2016 - 23:54
#2
Posted 13-07-2016 - 08:18
Cauchy Schwarz trực tiếp ta có: $P^2=(\sum \sqrt{\frac{2a}{b+c}})^2 \leq [\sum2b(a+c)][\sum \frac{1}{(a+c)(b+c)}]=\frac{8(ab+ac+bc)(a+b+c)}{(a+b)(b+c)(c+a)}$.
Dễ dàng chứng minh BĐT quen thuộc: $\frac{8(ab+ac+bc)(a+b+c)}{(a+b)(b+c)(c+a)} \leq 9$.
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