10 questions linked to/from Prove that $e^x\ge x+1$ for all real $x$
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### Simplest or nicest proof that $1+x \le e^x$

The elementary but very useful inequality that $1+x \le e^x$ for all real $x$ has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or ...
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### Proof of the inequality $e^x-x>0$ for all $x\in \Bbb R.$

Prove that the following inequality is true for all real numbers $$e^x-x>0.$$
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### Proof that $e^{-x} \ge 1-x$

My aim is to prove that $e^{-x} \geq 1-x$ for any $x \geq 0$. What I found so far is Bernoulli's inequality, which states that $$1+x\leq\left(1+\frac{x}{n}\right)^n\xrightarrow [n\to\infty]{} e^x$$ ...
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### For the function $y=\ln(x)/x$: Show that maximum value of y occurs when $x = e\ldots$

For the function $y=\ln(x)/x$: Show that maximum value of $y$ occurs when $x = e$. Using this information, show that $x^e <e^x$ for all positive values of $x$. Two positive integers, $a$ and $b$,...
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### Prove that $ex \leq e^x$ for all $x \in \mathbb{R}$

This is easy to prove for negative $x$ but what about positive $x$? Should I use MVT?
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### Prove that $e^x>x+1 \forall x\ne 0$ [duplicate]

I need to prove that $e^x>x+1 \forall x\ne 0$. Any ideas of hints about how to begin? I don't have any idea except the graphical way.
Let (*) be $\left(1+\frac{1}{n}\right)\left(1-\frac{1}{n^2}\right)^n < 1, ∀n∈ℕ$ I have tried many ways to get to (*) We have $\left(1-\frac{1}{n^2}\right)^n <1$ and $-\left(1+\frac{1}{n}\... 2answers 63 views ### Show that$\left(kx-1\right)e^{kx}>-1$holds$\forall k\in\mathbb{N}^*, \forall x\in\mathbb{R}^*$How to show that:$\forall k\in\mathbb{N}^*$and$\forall x\in\mathbb{R}^*$, the inequality$\left(kx-1\right)e^{kx}>-1$holds. Thank you for your help. 2answers 66 views ### How to prove that$er^2 \leq e^{r^2}$for all$r \in \mathbb{R}$I'm trying to prove the function$f:\mathbb{R}^2 \to \mathbb{R}\$ defined as $$f(x,y)=(x^2+y^2)e^{-(x^2+y^2)}$$ attains a maximum at every point of the unit circle. The determinant of the hessian ...
For any positive a and n, it seems this inequality holds $$\sum\limits_{t=n+1}^\infty e^{-at} \leq \frac{1}{a}e^{-an}$$ How can I prove this inequality and does this holds for negative a ?