Cho $a,b,c >0$ thỏa $ab+bc+ca=1$. Tìm GTNN của: $\sum \frac{\sqrt{a^2+1} \sqrt{b^2+1}}{ \sqrt{c^2+1}} $
Tìm GTNN
Started By dat102, 11-05-2018 - 13:36
#1
Posted 11-05-2018 - 13:36
$\sqrt{MF}$
#2
Posted 11-05-2018 - 13:56
Cho $a,b,c >0$ thỏa $ab+bc+ca=1$. Tìm GTNN của: $\sum \frac{\sqrt{a^2+1} \sqrt{b^2+1}}{ \sqrt{c^2+1}} $
Ta có: $\frac{\sqrt{a^2+1}\sqrt{b^2+1}}{\sqrt{c^2+1}}=\frac{\sqrt{a^2+ab+ac+bc}\sqrt{b^2+ab+ac+bc}}{\sqrt{c^2+ab+ac+bc}}=\frac{\sqrt{(a+b)(a+c)}\sqrt{(b+a)(b+c)}}{\sqrt{(a+c)(b+c)}}=a+b$
Tương tự cộng lại ta có: $P=2(a+b+c)\geq 2\sqrt{3(ab+bc+ca)}=2\sqrt{3}$
Dấu "=" xảy ra khi : $a=b=c=\frac{1}{\sqrt{3}}$
Edited by conankun, 11-05-2018 - 13:59.
- dat102, Khoa Linh and thanhdatqv2003 like this
$\large \mathbb{Conankun}$
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users