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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!

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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 $u\ge 0$.

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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.

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  • $\begingroup$ You're welcome, although I think that solution due to Michael Renardy is much more elegant. $\endgroup$ – Mateusz Wasilewski Jul 19 '12 at 15:02

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