Prove $F(x)=\frac{1}{x}\int_0^xf(t)$ is convex Assume $f(x)$ is convex in $[0,\infty)$, Prove
$F(x)=\frac{1}{x}\int_0^xf(t)dt$ is convex

0. As alluded to in the comments, there exists several characterizations of convex functions and each of them can prove useful in a given context: one can mention usual convexity inequalities, nondecreasing difference ratios, nondecreasing left- and right-derivatives, nonnnegative second derivatives (when these exist), the graph being above its tangents, some duality à la Fenchel-Legendre, etc. 
It happens that a simple change of variables allows to answer the specific question asked here rather shortly. Nevertheless we now explain another method, whose main advantage is that it allows to present some useful tools in this context.
To wit, a characterization of convexity is as follows: the function $f:[0,+\infty)\to\mathbb R$ is convex on $[0,+\infty)$ if and only if there exists a nondecreasing locally integrable function $g:(0,+\infty)\to\mathbb R$ and some finite $c$ such that, for every $x\geqslant0$,
$$
f(x)=c+\int_0^xg(y)\mathrm dy.
$$
(To be rigorous here, one can only be sure that $f(0)\geqslant c$, but to ask that $f$ is continuous at $0$ does not change its convexity.)
1. In our context, assuming that $f$ is convex and written as above, one gets
$$
F(x)=\frac1x\int_0^xf(y)\mathrm dy=c+\frac1x\int_0^x\int_0^yg(z)\mathrm dz\mathrm dy=c+\frac1x\int_0^xg(z)(x-z)\mathrm dz,
$$
and one seeks a nondecreasing locally integrable function $h$ such that $F=c+\int\limits_0^{\cdot} h$.
2. To find $h$, introduce the function 
$$
k=\int\limits_0^{\cdot}yg(y)\mathrm dy.
$$
An integration by parts yields
$$
F(x)=c+\int_0^xk'(z)\frac{x-z}{xz}\mathrm dz=c+\left[\frac{x-z}{xz}k(z)\right]_ 0^x+\int_0^xk(z)\frac{\mathrm dz}{z^2}.
$$
Since $k(z)=o(z)$ when $z\to0$, the second term on the RHS is zero and $F$ has the desired form, with
$$
h(x)=\frac1{x^2}k(x)=\frac1{x^2}\int_0^xyg(y)\mathrm dy.
$$
3. To check whether $h$ is nondecreasing, several methods are available here, let us explain one which is completely elementary. For every $0\lt x\leqslant y$,
$$
y^2h(y)=\int_0^xzg(z)\mathrm dz+\int_x^yzg(z)\mathrm dz=x^2h(x)+\int_x^yzg(z)\mathrm dz,
$$
and $g(z)\geqslant g(x)$ for every $z\geqslant x$, hence
$$
2y^2h(y)\geqslant 2x^2h(x)+g(x)\int_x^y2z\mathrm dz=2x^2h(x)+g(x)(y^2-x^2).
$$
Now, $g(z)\leqslant g(x)$ for every $z\leqslant x$ hence 
$$
2x^2h(x)\leqslant g(x)\int_0^x2z\mathrm dz=g(x)x^2,\quad\text{that is,}\quad g(x)\geqslant2h(x).
$$ 
Using this in the previous inequality, one gets 
$$
2y^2h(y)\geqslant 2x^2h(x)+2h(x)(y^2-x^2)=2y^2h(x),
$$
hence $h(y)\geqslant h(x)$ and the proof is complete.
