Bài toán 1 Cho $a+b+c=1$ và $a,b,c$ là các số thực dương. Chứng minh $\sqrt{a+b^{2}}+\sqrt{b+c^{2}}+\sqrt{c+a^{2}}\leq \frac{11}{5}$
Bài toán 2 ( dễ ) Cho $a,b,c> 0$ và $abc=1$. Chứng minh bất đẳng thức sau:
$\frac{3\sum a^{4}b^{4}}{a^{2}+b^{2}+c^{2}}+\frac{8a^{3}}{\left ( bc+a \right )^{3}}+\frac{8b^{3}}{\left ( ca+b \right )^{3}}+\frac{8c^{3}}{\left ( ab+c \right )^{3}}\geq 6$
$\sqrt{a+b^{2}}+\sqrt{b+c^{2}}+\sqrt{c+a^{2}}\leq \frac{11}{5}$
Started By tim1nuathatlac, 21-10-2012 - 18:47
#1
Posted 21-10-2012 - 18:47
#2
Posted 22-10-2012 - 20:56
Công nhận dễ thật !Bài toán 2 ( dễ ) Cho $a,b,c> 0$ và $abc=1$. Chứng minh bất đẳng thức sau:
$\frac{3\sum a^{4}b^{4}}{a^{2}+b^{2}+c^{2}}+\frac{8a^{3}}{\left ( bc+a \right )^{3}}+\frac{8b^{3}}{\left ( ca+b \right )^{3}}+\frac{8c^{3}}{\left ( ab+c \right )^{3}}\geq 6$
Theo AM-GM : $3(a^4b^4+b^4c^4+c^4a^4)\geq (a^2b^2+b^2c^2+c^2a^2)^{2}\geq 3a^2b^2c^2(a^2+b^2+c^2)= 3(a^2+b^2+c^2)$
$\Rightarrow \frac{3 (a^{4}b^{4}+b^{4}c^{4}+c^{4}a^{4})}{a^{2}+b^{2}+c^{2}}\geq 3$
Ta cần CM : $\frac{8a^{3}}{\left ( bc+a \right )^{3}}+\frac{8b^{3}}{\left ( ca+b \right )^{3}}+\frac{8c^{3}}{\left ( ab+c \right )^{3}}\geq 3$
$\Leftrightarrow \frac{1}{\left ( 1+\frac{bc}{a} \right )^{3}}+\frac{1}{\left ( 1+\frac{ca}{b} \right )^{3}}+\frac{1}{\left ( 1+\frac{ab}{c} \right )^{3}}\geq \frac{3}{8}$
Đặt $\frac{bc}{a}= x,\frac{ca}{b}= y,\frac{ab}{c}= z\Rightarrow xyz= 1$
$Q.E.D\Leftrightarrow \frac{1}{\left ( 1+x \right )^{3}}+\frac{1}{\left ( 1+y \right )^{3}}+\frac{1}{\left ( 1+z \right )^{3}}\geq \frac{3}{8}$
Quá quen rồi !
Edited by Secrets In Inequalities VP, 22-10-2012 - 20:57.
- donghaidhtt, WhjteShadow and chuot nhoc like this
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users