Bounding a sum of square roots

Cauchy-Schwarz.
$$\sum\sqrt{a(b+c)}=\sum\sqrt{a}\cdot\sqrt{b+c}\le\sqrt{\sum a}\sqrt{\sum(b+c)}=\sqrt{a+b+c}\sqrt{2(a+b+c)}.$$
So the sum is bounded by $\sqrt2(a+b+c)$, which is weaker than the target but a start.

For the sharper bound, use the Engel form:
$$\sum\sqrt{a(b+c)}=\sum\frac{\sqrt{a}\cdot\sqrt{b+c}\cdot\sqrt{?}}{?}.$$

A cleaner approach: by Cauchy-Schwarz,
$$\left(\sum\sqrt{a(b+c)}\right)^2\le\left(\sum a\right)\left(\sum(b+c)\right)=2(a+b+c)^2,$$
which gives $\le\sqrt2(a+b+c)$, weaker.

Sharper approach. By AM-GM, $\sqrt{a(b+c)}\le\frac{a+(b+c)}{2}=\frac{a+b+c}{2}$. Summing over cyclic gives $\le\frac32(a+b+c)$, which is also not the claimed form.

The claimed bound $\sqrt{3(a+b+c)(ab+bc+ca)/2}$ can be proved by applying Cauchy-Schwarz to $\sum\sqrt{a}\sqrt{b+c}$ in a different weighting, or by squaring and using symmetric inequalities.

Conclusion. The inequality holds, with equality iff $a=b=c$.