I have just seen a proof where they use this inequality, but I can’t figure out why is this true. I know that $e^x>x \forall x\in\mathbb{R}$ But I can’t get to this expression
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1$1+x\leq e^x$ for all real $x$ – Andrew Dec 08 '22 at 16:45
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1First, just replace $|x-y|$ by a variable $z\geq0,$ this asks if $e^{-z}\geq 1-z.$ – Thomas Andrews Dec 08 '22 at 16:46
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Hello, how can I prove that $1+x\leq e^x$ – RES Dec 08 '22 at 16:48
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It isn't 100% obvious that $e^{-z}\geq 1-z,$ but it is obvious if $z\geq 1,$ and if $0\leq z<1,$ then $$e^{-z}=1-z+z^2/2-z^3/3!+\cdots$$ is an alternating decreasing sequence, and so we get $1-z\leq e^{-z}.$ – Thomas Andrews Dec 08 '22 at 16:51
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You can differentiate $f(x)=e^x-x-1$, use taylor expansion of $e^{x}$, etc.. – Hypernova Dec 08 '22 at 16:51
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Or, for $0\leq z<1,$ this is equivalent to: $e^{z}\leq \frac1{1-z},$ which we can see just by comparing the power series. – Thomas Andrews Dec 08 '22 at 16:53
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Another way to prove $e^w\geq 1+w$ for all real $w$ is to use Bernoulli's inequality, which states that if $u\geq -1$ and $n$ is a positive integer, then $(1+u)^n\geq 1+nu.$ Then use that: $(1+w/n)^n\to e^w,$ and, for large $n,$ $w/n\geq -1,$ so $(1+w/n)^n\geq 1+w.$ – Thomas Andrews Dec 08 '22 at 16:59
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Let $h:[0,\infty)\to\mathbb{R}$ , $h(x)=e^x-1-x$ then $h’(x)=e^x-1 \geq 0$ so h is increasing. Even more, $h(x)\geq h(0)=0$ So if $x\geq 0$, then $h(x)\geq 0\Rightarrow e^x\geq 1+x$
Thomas Andrews
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EMS
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You've only proven $e^x\geq 1+x$ for $x\geq0,$ but we specifically need it proven for $x<0.$ – Thomas Andrews Dec 08 '22 at 17:02
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You can adjust this proof by showing that $h'(x)$ means $h$ is decreasing for $x<0$ and increasing after, so $h(0)=0$ is the minimum. – Thomas Andrews Dec 08 '22 at 17:04
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Since the exponential is a convex function, it is above its tangent at $x=0$, that is $$ e^x \geq 1 + x $$
LL 3.14
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