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We know that $x_1\lambda_1=x_2\lambda_2$ and $x_1,x_2,\lambda_1,\lambda_2>0$. We also know that $x_1\lambda_1>1$ and $x_2\lambda_2>1$. We also have $\lambda_1 > \lambda_2$. Now we want to prove the following:

$$\frac{\lambda_1}{e^{\lambda_2x_1}}-\frac{\lambda_2}{e^{\lambda_1x_2}}\le\lambda_1-\lambda_2$$

I've done some simulation in MATLAB and I am strongly convinced that this can be proved. The only question is that I don't know how.

I am willing to provide further details if you need.

Thanks.

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    I'm not sure this can be helpful, but simplification will give the following $\frac{1}{x_{1}} \frac{1}{e^{a \frac{x_{1}}{x_{2}}}} - \frac{1}{x_{2}} \frac{1}{e^{a \frac{x_{2}}{x_{1}}}} \leq \frac{1}{x_{1}} - \frac{1}{x_{2}}$ – 2017-02-12
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    Sorry but I don't quite get the meaning of $a$ in your comments, would you please explain a bit more? – 2017-02-12
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    $a= \lambda_{1}x_{1} = \lambda_{2}x_{2}$. – 2017-02-12

1 Answers 1

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The inequality does not hold in general. For $\lambda_1 = 2$, $x_1 = 0.55$, $\lambda_2 = 1$, $x_2 = 1.1$ $$ \frac{\lambda_1}{e^{\lambda_2x_1}}-\frac{\lambda_2}{e^{\lambda_1x_2}} \approx 1.043 > 1 = \lambda_1-\lambda_2 $$

It seems to be wrong for arbitrary $\lambda_1 > \lambda_2$ if $x_1\lambda_1=x_2\lambda_2$ is close enough to $1$.

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    Thanks very much for the answer! Is there any possibility to make this inequality hold by adding extra conditions. Such as we require $x_1\lambda_1 > 1+\delta$ for some $\delta > 0$ – 2017-02-12
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    @YiDing: That may be but I do not have an answer for that at present. – 2017-02-12
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    Just for fun. Maybe we can take $1+\delta = 1,176280818...$. (It's Mahler measure(hieght) of Lehmer's polynomial.). – 2017-02-12