Hello everyone, could someone gives the conditions for $\lambda$ that the following inequality is correct for any $0\leq x\leq\alpha$ 


$$\lambda x-1+e^{-\lambda x}\leq\lambda^2x\sqrt{\alpha x}$$


Here we know $2\leq\alpha\leq 4$ and $\lambda\alpha\geq 1$.


Thanks for help!



inequality of a function
It is true for any positive $\lambda$.
If $x\le\alpha$, the right hand side is greater or equal to $\lambda^2x^2$. Hence it suffices to show that $u-1+\exp(-u)-u^2\le 0$ for positive $u$. It is easy to show that the function
$f(u)=u-1+\exp(-u)-u^2$ is monotone decreasing and concave for $u\ge 0$.


