How to show that:
$\forall k\in\mathbb{N}^*$ and $\forall x\in\mathbb{R}^*$, the inequality $\left(kx-1\right)e^{kx}>-1$ holds.
Thank you for your help.
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1It's just $e^y>1+y$ for $y=-kx$. – yurnero Jan 15 '17 at 15:29
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Alright :D, thanks. – PMC1234 Jan 15 '17 at 15:33
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See: http://math.stackexchange.com/questions/252541/prove-that-ex-ge-x1-for-all-real-x – yurnero Jan 15 '17 at 15:34
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Yes it's with Bernoulli then... Thank you for your help @yurnero – PMC1234 Jan 15 '17 at 15:38
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It results from the variations of $f$ on $\mathbf R$: $f'(x)=k^2x\mathrm e^{kx}$, so $f$ decreases on $\mathbf R^-$, increases on $\mathbf R^+$ and has a minimum at $x=0$, hence for $x\ne 0$, $f(x)>f(0)=-1$.
Bernard
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I think $x=0$ or $k=0$ is ruled out by the specification of the problem. – yurnero Jan 15 '17 at 15:36
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We consider the function $f$ defined on $\mathbb{R}$ such that $f(x)=(kx-1)e^{kx}$ with $k$ a natural integer (excluding zero).
Let $X=kx$. We then have $f(X)=(X-1)e^X$. Let's show that $f$ is strictly superior to $-1$.
$f$ is differentiable on $\mathbb{R}$ where: \begin{align*} \forall X\in\mathbb{R},\ f'(X)&=\left((X-1)e^X\right)'\\ &=Xe^X \end{align*} Knowing that $e^X>0$ on $\mathbb{R}$, then $f'$ is of the sign of $X$. $f$ is strictly decreasing for $X<0$ and strictly increasing for $X>0$. Hence, the function has a minimum at $X=0$ which equals: \begin{align*} f(0)&=(0-1)e^0\\ &=-1 \end{align*} Thus, if we exclude X=0, we have a fortiori, $f(X)>f(0)\Leftrightarrow f(x)>-1\Leftrightarrow (kx-1)e^{kx}>-1$.