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I am trying to prove that

$$e^{k+1} ≥ 3 + 3k + k^2$$ with, $$k>2$$

WhatI have done so far:

What we are trying to prove is that $$e^n≥1+n+n^2$$ is a true statement. Since $n=3$ holds, this is our base case. Then, the inductive hypothesis is: $$e^k≥1+k+k^2$$ Hence, I need to show $$e^{k+1}≥1+(k+1)+(k+1)^2=3+3k+3k^2$$ is a true statement.

Lerbi
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  • For the inductive step we get: $$e^{k+1}\geq e(1+k+k^2) $$ By the inductive hypothesis. Now you want to solve $e(1+k+k^2)\geq 3+3k+k^2 $ or $(e-1)k^2+(e-3)k+(e-3)\geq 0 $ and show this holds for all $k \geq 3$ – Raxel Nov 21 '14 at 07:33

2 Answers2

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I think you should try using Taylor series of $e^{x}$ here.

wirm
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From a purely algebraic point of view, let us consider the function $$f(k)=e^{k+1}-( 3 + 3k + k^2)$$ Its derivative $$f'(k)=e^{k+1}-2 k-3=0$$ has an analytical solution $$k=-W_{-1}\left(-\frac{1}{2 \sqrt{e}}\right)-\frac{3}{2}\approx 0.256431$$ where appears Lambert function. At this point $$f(k)=-\frac{1}{4} \left(2 W_{-1}\left(-\frac{1}{2 \sqrt{e}}\right)+1\right) \left(2 W_{-1}\left(-\frac{1}{2 \sqrt{e}}\right)+3\right)\approx -0.322188$$ The second derivative test shows that this is a minimum.

Moreover, $f(1)=e^2-7>0$ then the inequality holds for any $k>1$.

Claude Leibovici
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