Does $f'(x)>\frac{f(x)}{x}$ for $x>0$ imply $f$ is convex?

Let us consider a simple polynomial function $f(x) = ax^4+bx^3+cx^2$. Obviously we have $f(0)=0$, to satisfy the condition, we need \begin{equation} f'(x) = 4ax^3+3bx^2+2cx > \frac{f(x)}{x} = ax^3+bx^2+cx. \end{equation} Thus $3ax^2+2bx + c > 0$ will be held for any $x$, i.e., \begin{equation} a>0, \quad b^2 - 3ac < 0. \end{equation} To violate the convexity, we need some $x_0 > 0$, such that \begin{equation} f''(x_0) = 12a x_0^2+6b x_0 + 2c < 0. \end{equation} Thus what we need is \begin{equation} \left\{ \begin{array}{c} {a>0,} \\ {b^2 - 3ac < 0,} \\ {\exists x_0, 12ax_0^2 + 6bx_0 + 2c <0.} \end{array} \right. \end{equation}

Here we simply choose $x_0=1$, and we can easily find a feasible solution: \begin{equation} a=0.1, \quad b=-0.4, \quad c=0.55. \end{equation}

And I plot a simple figure in the following: