Let $x>0$. Prove that: $$x^x-1\geq(x-1)e^{x-1}$$
I tried to use the Taylor series, but I did not get a proof.
Prove that $x^x-1\geq(x-1)e^{x-1}$
2
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calculus
real-analysis
inequality
1 Answers
7
If we prove that $f(x)=(x^x-1)e^{1-x}$ is a convex function on $\mathbb{R}^+$ we are done, since $y=x-1$ is just the equation of the tangent at $x=1$. On the other hand $$ f''(x)=\frac{e^{1-x}}{x}\left(-x+x^x\left(1+x\log^2 x\right)\right)\tag{1} $$ and $(1+x\log^2 x)\geq 1$, $x^x\geq x$, so $f$ is trivially convex on $\mathbb{R}^+$. Just to clarify: $x^x\geq x$ is equivalent to $\exp\left((x-1)\log x\right)\geq 1$, that is trivial too, since $x-1$ and $\log x$ have the same sign on $\mathbb{R}^+$ and $\exp$ is monotonic.