We want to prove that $f(x) := \lambda^{2} x \sqrt{\alpha x} - \lambda x + 1 - e^{-\lambda x}$ is nonnegative for $x \in [0,\alpha]$. Assume that $f(\xi)=0$ for some $\xi \in (0, \alpha]$; we shall show that the derivative is positive at this point. Since $f(\xi)=0$, we may write
$$ e^{-\lambda \xi} -1 = \lambda^2 \sqrt{\alpha} \xi^{\frac{3}{2}} - \lambda \xi.$$
We plug it into the formula for derivative and obtain
$$f'(\xi) = \frac{3}{2} \lambda^{2} \sqrt{\alpha} \sqrt{\xi} + \lambda^{3} \sqrt{\alpha} \xi^{\frac{3}{2}} - \lambda^{2} \xi .$$
Let's consider new variable $t = \left(\frac{\xi}{\alpha}\right)^{\frac{1}{2}}$, then one may rewrite the above formula in following manner:
$$f'(\xi) = \frac{3}{2} \lambda^2 \alpha t + \lambda^{3} \alpha^{2} t^{3} - \lambda^{2} \alpha t^{2};$$
we can divide by $\lambda^{2} \alpha$, as it is some positive factor. Our task now is to prove that the polynomial $p(t) = \lambda\alpha t^{3} - t^{2} + \frac{3}{2} t$ is positive on $(0,1]$. But its derivative is equal to $p'(t) = 3 \lambda \alpha t^2 - 2t + \frac{3}{2}$ and is positive, due to condition $\lambda \alpha \geqslant 1$.

I hope there aren't any errors.