Prove that $a \sqrt{b + c} + b \sqrt{c + a} + c \sqrt{a + b} \le \sqrt{2(a+b+c)(bc + ac + ab)}$ for $a, b, c > 0$ Prove for $a, b, c > 0$ that 
$$a \sqrt{b + c} + b \sqrt{c + a} + c \sqrt{a + b} \le \sqrt{2(a+b+c)(bc + ac + ab)}$$
Could you give me some hints on this?
I thought that Jensen's inequality might be of use in this exercise, but I haven't managed to solve this on my own.
 A: Use the Cauchy-Schwarz inequality on the two vectors $(\sqrt a, \sqrt b, \sqrt c)$ and $(\sqrt{a(b+c)}, \sqrt{b(a+c)}, \sqrt{c(a+b)})$ (then take the square root on both sides or not, depending on which version of the CS inequality you use), and lastly note that we have:
$$
 a(b+c) + b(a + c) + c(a + b) = 2(bc + ac + ab)
$$
A: If you look at C-S for just two terms and its use here becomes more obvious
$$ (a_1b_1 + a_2b_2)^2 \le (a_1^2+a_2^2)(b_1^2+b_2^2) $$
Now look at inequality we want to prove. Square both sides and examine how the terms match with our C-S
$$(a\sqrt{b+c} + b\sqrt{a+c} + c\sqrt{a+b})^2 \le 2(a+b+c)(ab+ac+bc)$$
We can see how our inequality will resemble C-S by taking vectors
$$a_i = (\sqrt{a},\sqrt{b},\sqrt{c})$$
$$b_i = (\sqrt{a(b+c)}, \sqrt{b(a+c)}, \sqrt{c(a+b)} )$$
By applying C-S to the above vectors we aim to prove our original inequality
$$(a\sqrt{b+c} + b\sqrt{a+c} + c\sqrt{a+b})^2 \le a(b+c) + b(a+c) +c(a+b)$$
Simplifying RHS
$$(a\sqrt{b+c} + b\sqrt{a+c} + c\sqrt{a+b})^2 \le2(ab+ac+bc)$$
Take square root of both sides
$$a\sqrt{b+c} + b\sqrt{a+c} + c\sqrt{a+b} \le \sqrt{2(ab+ac+bc)}$$
A: Another way:
By AM-GM we obtain:
$$\sum_{cyc}a\sqrt{b+c}=\sqrt{\sum_{cyc}(a^2b+a^2c+2ab\sqrt{(b+c)(a+c)}}\leq$$
$$\leq\sqrt{\sum_{cyc}(a^2b+a^2c+ab(b+c+a+c)}=\sqrt{2(a+b+c)(ab+ac+bc)}.$$
