I have a differential inequality as follows: $$f'(x)\geq cf(x)^a,\quad \forall x\in[0,1]$$ where $0<a<1,\, f(x)\geq 0$
Wolfram Alpha gives the following answer for the equality: $$f(x)= ((a - 1) (k_1 - c x))^{1/(1 - a)}$$
I'm interested to find the answer to the inequality. But I don't know where to begin.
On
If $f(x)=0$ for some value of $x$, then $f'(x)$ must be positive or zero to maintain the inequality $f(x)\ge 0$. If we assume a smooth function $f(x)$, $f'(x)\ge0$ would imply that immediately prior to $f(x)=0$ we had $f(x)<0$, and so if $f(x)=0$ we require $f'(x)=0$. This yields the trivial solution that $f(x)=0$ for all $x$.
Now consider $f(x)>0$. In this case we may write \begin{equation}\tag{1} \frac{df}{dx}\ge cf^a \Rightarrow \frac{df}{f^a}\ge c\ dx. \end{equation} This is true for all values of $f(x)$ and $x$, and so integrating maintains the inequality: \begin{equation}\tag{2} \frac{f(x)^{1-a}}{1-a}\ge cx +d, \end{equation} where $d$ is some constant. Since $0<a<1$ we find that $1-a>0$, and so \begin{equation}\tag{3} f(x)^{1-a}\ge (1-a)(cx+d). \end{equation} Finally, since exponentiation by a real positive quantity is monotonic we conclude \begin{equation}\tag{4} f(x) \ge \left[(1-a)(cx+d)\right]^{1/(1-a)}. \end{equation}
Introduce a new function $r(x)\geq 0$ and add $r(x)f(x)$ to the right side of the inequalty to make it an equality.
$$f'(x)=r(x)f(x)+cf(x)^a$$
Now, solve this Bernoulli differential equation by the substitution
$$z(x)=f(x)^{1-a}.$$
After having obtained the final solution try to eliminate $r(x)$ by estimating the solution.