Cho $a,b,c>0$ và $a+b+c= \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Chứng minh rằng
$3(a+b+c)\geq \sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}$
Cho $a,b,c>0$ và $a+b+c= \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Chứng minh rằng
$3(a+b+c)\geq \sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}$
Cho $a,b,c>0$ và $a+b+c= \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Chứng minh rằng
$3(a+b+c)\geq \sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}$
Solution: Ap dụng bđt Cauchy-Schwarzt thì ta có:
$(\sqrt{8a^{2}+1}+\sqrt{8b^{2}+1}+\sqrt{8c^{2}+1})^{2}\leq (a+b+c)(\frac{8a^{2}+1}{a}+\frac{8b^{2}+1}{b}+\frac{8c^{2}+1}{c})$
Mà $ (a+b+c)(\frac{8a^{2}+1}{a}+\frac{8b^{2}+1}{b}+\frac{8c^{2}+1}{c})= (a+b+c)(8a+8b+8c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
$= (a+b+c)(8a+8b+8c+a+b+c)= 9(a+b+c)^{2}$
Suy ra $(\sqrt{8a^{2}+1}+\sqrt{8b^{2}+1}+\sqrt{8c^{2}+1})^{2}\leq 9(a+b+c)^{2}$
$\Leftrightarrow \sqrt{8a^{2}+1}+\sqrt{8b^{2}+1}+\sqrt{8c^{2}+1}\leq 3(a+b+c)$ Q.E.D
DBRX khi $a=b=c=1$
Edited by NHoang1608, 09-03-2017 - 21:23.
The greatest danger for most of us is not that our aim is too high and we miss it, but that it is too low and we reach it.
----- Michelangelo----
Edited by Kamii0909, 10-03-2017 - 17:41.
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