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Let $0<b_1<b_2$ Let $-\infty<x_1<x_2<\infty$. Suppose $a_1,a_2$ satisfies the simultaneous equations $$a_1e^{b_1x_1}+a_2e^{b_2x_1}=a_1e^{b_1x_2}+a_2e^{b_2x_2}=1$$ Then show that $f(x)=a_1e^{b_1x}+a_2e^{b_2x} > 1$ iff $x_1<x<x_2$.

I checked for few values in my computer. The result holds good. I was able to show that $a_1>0$ and $a_2<0$. Plugging the values of $a_1,a_2$ gives us a nasty function. I wonder if there is an easy way to show it.

Any suggestions/help.

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    $\begingroup$ I am sorry but your title (concavity of $f$) is not equivalent to property $f(x)=a_1e^{b_1x}+a_2e^{b_2x} > 1$ $\endgroup$
    – Jean Marie
    Nov 16 '16 at 22:07
  • $\begingroup$ @JeanMarie Concavity is an additional thing that may not be true (and hence a question mark).. It is my hunch. However that result is supposed to be true $\endgroup$
    – Sayan
    Nov 16 '16 at 22:15
  • $\begingroup$ Advice: why not ask 2 questions a) show that $f$ is .... >1 b) Can we deduce that $f$ is concave ? Moreover, include condition $a_2<0<a_1$. $\endgroup$
    – Jean Marie
    Nov 16 '16 at 22:26
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HINT.- Making $$\begin{cases}x=\lambda x_1+(1-\lambda)x_2;\space 0\lt\lambda\lt1\\0\lt b_1=\mu b_2; \ 0\lt\mu\lt1\\ e^{b_2x_1}=t\text{ and }\space e^{b_1x_2}=s\end{cases}$$ one has

$$f(x)=a_1t^{\mu\lambda}s^{\mu(1-\lambda)}+a_2t^{\lambda}s^{1-\lambda}=a_1\left(\frac ts\right)^{\mu\lambda}s^{\mu}+a_2\left(\frac ts\right)^{\lambda}s$$

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  • $\begingroup$ Thanks. $s=e^{b_2x_2}$ I guess. However I cant see how it helps. Can you elaborate? $\endgroup$
    – Sayan
    Nov 17 '16 at 11:06

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