Solution:
Applying $x+y \leq \sqrt{2\left(x^{2}+y^{2}\right)}$ for $x=\sqrt{\frac{a^{2}+a b+b^{2}}{3}}$ and $y=\sqrt{a b}$, we will obtain
$$\sqrt{\frac{a^{2}+a b+b^{2}}{3}}+\sqrt{a b} \leq \sqrt{\frac{2 a^{2}+2 a b+2 b^{2}+6 a b}{3}} \leq \sqrt{\frac{3\left(a^{2}+b^{2}+2 a b\right)}{3}}=a+b.$$

The inequality is equivalent to
$$\frac{a^{2}+a b+b^{2}}{3}+\frac{3 a b}{3}+2 \sqrt{\frac{a b\left(a^{2}+a b+b^{2}\right)}{3}} \leq \frac{3 a^{2}+6 a b+3 b^{2}}{3}.$$
This can be rewritten as
$$2 \sqrt{\frac{a b\left(a^{2}+a b+b^{2}\right)}{3}} \leq \frac{2\left(a^{2}+a b+b^{2}\right)}{3}$$
or
$$\sqrt{a b} \leq \sqrt{\frac{a^{2}+a b+b^{2}}{3}}$$
which is obviously true since $a^{2}+b^{2}+a b \geq 2 a b+a b=3 a b$.