<s>[INST] Simple inequality with a,b,c I'm looking for proof of
$$\sqrt{a(b+c)}+\sqrt{b(c+a)}+\sqrt{c(a+b)} \leq \sqrt{2}(a+b+c)$$
I tried using $m_g \leq m_a$, generating the permutations of $\sqrt{a(b+c)} \leq \frac{a+(b+c)}{2}$ and adding them together, but I get $\frac{3}{2}(a+b+c)$, and obviously $\frac{3}{2}>\sqrt{2}$, so I'm stuck and sure I'm missing something trivial, so thanks in advance.
 [/INST] Using Cauchy-Schwarz inequality with the vectors:
$$\begin{cases}
u=(\sqrt{a},\sqrt{b},\sqrt{c})\\
v=(\sqrt{b+c},\sqrt{c+a},\sqrt{a+b})
\end{cases}$$ you get
\begin{align}[\sqrt{a(b+c)}+\sqrt{b(c+a)}+\sqrt{c(a+b)}]^2 &=(u.v)^2\\ &
\le \Vert u \Vert^2 \Vert v \Vert^2\\
&=(a+b+c)((b+c)+(c+a)+(a+b))\\
&=2(a+b+c)^2\end{align} Which enables to conclude to the desired result.
</s>