A direct application of Hölder

Cauchy–Schwarz. For the two vectors $u=\bigl(a^{3/2},b^{3/2},c^{3/2}\bigr)$ and $w=\bigl(a^{1/2},b^{1/2},c^{1/2}\bigr)$,
$$\langle u,w\rangle^2\le\|u\|^2\,\|w\|^2.$$
Now $\langle u,w\rangle=\sum a^{3/2}\cdot a^{1/2}=\sum a^2=a^2+b^2+c^2$, while $\|u\|^2=\sum a^3$ and $\|w\|^2=\sum a=a+b+c$. Hence
$$(a^2+b^2+c^2)^2\le(a^3+b^3+c^3)(a+b+c).$$

Equality. Equality in Cauchy–Schwarz holds iff $u$ and $w$ are proportional. For $a,b,c>0$, this means $a^{3/2}/a^{1/2}=a$ is constant, i.e. $a=b=c$. (If some variables vanish, equality holds iff the nonzero ones are equal.)

Remark. The inequality is a special case of Hölder's inequality: for $p=3/2$ and $q=3$, applied to the products $a^{3/2}\cdot a^{1/2}$. It is also the $p=3$, $r=1$ case of the power-mean inequality $\bigl(\frac{a^r+b^r+c^r}{3}\bigr)^{1/r}\le\bigl(\frac{a^s+b^s+c^s}{3}\bigr)^{1/s}$ for $r\le s$.

Conclusion. $(a^3+b^3+c^3)(a+b+c)\ge(a^2+b^2+c^2)^2$, with equality iff $a=b=c$ (for positive $a,b,c$).