About an integral inequality Let $ u$ be a nonnegative continous function defined in $[x_0, x_0+a ] $, and $\alpha \in(0,1)$.  
Prove that if $$u(x) \le \int\limits_{x_0}^x \! Cu^\alpha(t) dt$$
with $C\ge 0$ then
$$u(x) \le (C(1-\alpha)(x-x_0))^{1/(1-\alpha)} .$$
 A: Assume without loss of generality that $x_0 = 0$. We can also 
assume that $u$ is not identically zero and
$$
    x_1 = \inf \{ x \in [0, a] \mid u(x) > 0 \}  = 0
$$
Otherwise do the following estimate on the interval $[x_1, a]$, this
will only strengthen the inequality.
Define $v(x)$ for
 $0 \le x \le a$ as
$$
   v(x) = \left( \int_0^x u^\alpha(t) \, dt \right)^{1-\alpha} 
$$ 
Then $v(0) = 0$. $v$ is positive and differentiable on the interval $(0, a)$, with
$$
 v'(x) = (1-\alpha) \left( \int_0^x u^\alpha(t) \, dt \right)^{-\alpha}
  u^\alpha(x) \\
 \le (1-\alpha) \left( \int_0^x u^\alpha(t) \, dt \right)^{-\alpha}
  C^\alpha \left( \int_0^x u^\alpha(t) \, dt \right)^{\alpha}
 = (1-\alpha) C
$$
It follows that for $x \ge 0$
$$
 v(x) \le (1-\alpha) C x
$$
and therefore
$$
 u(x) \le C \int_0^x u^\alpha(t) \, dt
 = C v(x)^{1/(1-\alpha)} \le (C(1-\alpha)(x-x_0))^{1/(1-\alpha)}
$$
A: $\textbf{WRONG ANSWER}$ (I cannot assert the first inequality in terms of derivatives)
Derive the expression with respect to $x$ applying Leibniz's rule resulting 
$$u'(x)\leq Cu^{\alpha}(x)$$
This is an easy diff. equation
$$\frac{u'(x)}{u^{\alpha}(x)}\leq C$$
Integrating in the interval yields to the desired expression.
