Because $a, b, c$ are side lengths of a triangle, we have $a+b-c>0, b+c-a>0,$ and $c+a-b>0$. Therefore, we can write \begin{align} \sqrt{a}(c+a-b)&=\sqrt{a(c+a-b)}\sqrt{c+a-b}\\ \sqrt{b}(a+b-c)&=\sqrt{b(a+b-c)}\sqrt{a+b-c}\\ \sqrt{c}(b+c-a)&=\sqrt{c(b+c-a)}\sqrt{b+c-a} \end{align} Apply Cauchy inequality (see %%HREF%%3865%%) on $\vec{u}=(\sqrt{a(c+a-b)}, \sqrt{b(a+b-c)},\sqrt{c(b+c-a)})$ and $\vec{v}=(\sqrt{c+a-b},\sqrt{a+b-c},\sqrt{b+c-a})$ yields: \begin{align} &(\sqrt{a}(c+a-b) + \sqrt{b}(a+b-c)+\sqrt{c}(b+c-a))^2\\ & \le (a(c+a-b) + b(a+b-c) + c(b+c-a))((c+a-b)+(a+b-c)+(b+c-a))\\ &= (a^2+b^2+c^2)(a+b+c) \end{align} Taking square root of both sides leads to the desired result.