An inequality

Yes, with equality exactly when $a=b=c=\tfrac19$.
Set $x=\sqrt a$, $y=\sqrt b$, $z=\sqrt c$, so that $x,y,z>0$ and $x+y+z=1$. Since
$a+b+c=x^{2}+y^{2}+z^{2}$ and $a^{3}+b^{3}+c^{3}=x^{6}+y^{6}+z^{6}$, the claim reads
$$\bigl(x^{2}+y^{2}+z^{2}\bigr)^{5}\ \leqslant\ x^{6}+y^{6}+z^{6}. \qquad \tag{$\ast$}$$The constraint $x+y+z=1$ with $x,y,z>0$ says precisely that $(x,y,z)$ is a probability distribution. Let $X$ be the random variable taking the value $x$ with probability $x$, the value $y$ with probability $y$, and the value $z$ with probability $z$. Then
$$\mathbb{E}[X]=x^{2}+y^{2}+z^{2},\qquad \mathbb{E}\bigl[X^{5}\bigr]=x^{6}+y^{6}+z^{6}.$$The map $u\mapsto u^{5}$ is convex on $\mathbb{R}_{+}$, so Jensen’s inequality gives
$$\mathbb{E}[X]^{5}\ \leqslant\ \mathbb{E}\bigl[X^{5}\bigr],$$which is exactly $(\ast)$. $\blacksquare$

on applique l’inégalité de Ho''lder avec $p=5$ et $q=5/4$ (on a bien $1/p+1/q=1$) et en écrivant $x=(x^{3})^{1/5}(\sqrt x)^{4/5}$ pour a, b et c