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How to prove $$\displaystyle \frac{{b{e^{bx}} - a{e^{ax}}}}{{b - a}}<e^{(a+b)x}$$ for $0<a<b$ and $x\ne 0$ ?

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    $\begingroup$ So in other words you want to prove that $be^{bx}\lt (b-a)e^{(a+b)x}+ae^{ax}$. What have you tried so far? $\endgroup$ – abiessu Oct 3 '13 at 18:08
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As mentioned in the comments, we can rearrange the original inequalities as follows: $$ \displaystyle {{b{e^{bx}} }}<(b-a)e^{(a+b)x}+a{e^{ax}} $$ Here you can use Jensen but if you do not want to use notion of convexity, you can use the weighted arithmetic geometric inequality, which is: $$ w_1x_1+w_2x_2\geq (w_1+w_2)(x_1^{w_1}x_2^{w_2})^{\frac{1}{w_1+w_2}} $$ which gives you: $$ \displaystyle (b-a)e^{(a+b)x}+a{e^{ax}}\geq (b-a+a)(e^{a.ax+(b-a)(b+a)x})^{\frac{1}{b}} $$


This does NOT mean that the proof is independent of Jensen inequality. It is only so, if you do not use Jensen inequality to prove weighted AG inequality.

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Hint: Use Jensen inequality for $e^x$ with suitable weights...

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The proof independent of Jensen's: multiply both sides by the denominators and group the terms to end up with
$$\frac{be^{bx}}{e^{bx}-1}\ge \frac{ae^{ax}}{e^{ax}-1}.$$ Now fix $x$ and consider the function $f(t)=\frac{te^{tx}}{e^{tx}-1}=t+\frac{t}{e^{tx}-1}.$ Note, $f'(t)=\frac{e^{tx}(e^{tx}-1-tx)}{(e^{tx}-1)^2}\ge 0,$ since $e^y\ge 1+y$ and thus $f(b)\ge f(a)$ for $b\ge a>0.$

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  • $\begingroup$ first of all $e^y\ge 1+y$ for ALL $y.$ Second of all the derivative I found is correct modulo that i did not close the bracket for the power which of course does not affect the solution. $\endgroup$ – leshik Oct 3 '13 at 20:13
  • $\begingroup$ And the final note: this method should work as long as the problem statement is correct, since all over conclusions work in two directions (if and only if). $\endgroup$ – leshik Oct 3 '13 at 20:19
  • $\begingroup$ sorry! you are right. $\endgroup$ – mert Oct 4 '13 at 6:41

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