Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question

2 Answers 2

up vote 2 down vote accepted

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 $u\ge 0$.

share|improve this answer
    
perfect! Thank you very much –  Higgs88 Jul 19 '12 at 20:57

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.

share|improve this answer
    
excellent reply! Thanks a lot –  Higgs88 Jul 19 '12 at 14:53
    
You're welcome, although I think that solution due to Michael Renardy is much more elegant. –  Mateusz Wasilewski Jul 19 '12 at 15:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.