# Tagged Questions

61 views

### Solve the inequality $(1/2)^x-(1/2)^{-1-x}\ge1$ for real $x$

I have to solve in $\Bbb{R}$ the following inequality : $$\left(\frac{1}{2}\right)^{x} - \left(\frac{1}{2}\right)^{-1 - x} \ge 1 \qquad(E)$$ So far I have : For $x=0$ this inequality if not ...
143 views

### What $\alpha$ such that if $xy=\alpha$, then $e^{-x}+e^{-y}\geq 2e^{-\sqrt \alpha}$?

For every $x,y \gt 0$, if $xy=\alpha$, then we have $$e^{-x}+e^{-y}\geq 2e^{-\sqrt \alpha}$$ What are the possible values of $\alpha$? $2 < e^{1/(n+1)} + e^{-1/n}$ led to this problem. ...
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### Is $\int_x^{\infty}e^{-\frac{t^2}{2}} < \frac{1}{x}e^{-\frac{x^2}{2}}$?

While solving a problem in real analysis, I got stuck. I need to prove $$\int_x^{\infty}e^{-\frac{t^2}{2}}dt < \frac{1}{x}e^{-\frac{x^2}{2}}$$ Clearly I have to use some kind of inequality, but ...
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### An exponential/polynomial inequality

Prove that there is at least $1$ real number $a>0$ with the property $$a^x\ge x^a$$ for any $x>0$.
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### Showing $n!<e(\frac{n}{2})^n$

I'd like to prove that $n!<e(\frac{n}{2})^n$. What I have so far: $\sqrt[n]{n!} = \sqrt[n]{1\cdot 2 \cdot \ldots \cdot n} \leq \frac{1+\ldots +n}{n}=\frac{(n+1)n}{2n}=\frac{(n+1)}{2}$. Thus ...
710 views

### How to solve exponential function inequality?

How do I solve the exponential equations like $2^\frac{x}{8}<x$? I can solve this by plotting into graph. But is there any way to do it mathematically? or like $2^x < 100x^2$ . I am trying to ...
305 views

### proof of inequality $e^x\le x+e^{x^2}$
Does anybody have a simple proof this inequality $$e^x\le x+e^{x^2}.$$ Thanks.