About an integral inequality

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 $$ u(x) \le C \int_0^x u^\alpha(t) \, dt = C v(x)^{1/(1-\alpha)} \le (C(1-\alpha)(x-x_0))^{1/(1-\alpha)} $$