Prove $e^2\ge2^e$.
I've thought of applying $\ln$ in both sides and that's what I get: $$e^2\ge2^e \Leftrightarrow \ln e^2\ge \ln 2^e \Leftrightarrow \ln e^2-\ln 2^e\ge 0\Leftrightarrow \ln(e^2/2^e)\ge0 \Leftrightarrow e^{\ln(e^2/2^e)}\ge e^0 \Leftrightarrow \dfrac {e^2}{2^e}\ge 0 (1) $$ Then I multiplied both sides with $$2^e$$ So ($1$) becomes: $$e^2\ge 0 $$ which is true for $$\forall x\in \mathbb R$$Is this the correct way of proving that $$e^2\ge 2^e$$ is true?
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taking the logarithm on both sides we obtain $$\frac{2}{\ln(2)}\geq e$$ and now consider the function $$\frac{x}{\ln(x)}$$ we get $$f'(x)$$ by the Quotient rule: $$f'(x)=\frac{\ln(x)-1}{\ln(x)^2}$$ and $$f'(x)=0$$ if $$x=e$$
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Note that with your work you arrived to
$$...e^{ln(e^2/2^e)}\ge e^0 \iff e^2/2^e\ge \color{red}1$$
and not zero.
Note also that at the same inequality we can arrive directly by
$$e^2\ge 2^e \iff \frac{e^2}{2^e}\ge 1$$
the best way is to take log both sides then
$$e^2\ge 2^e \iff 2\log e\ge e\log 2 \iff 2 \ge e\log 2$$
which is easy to check.
On
Comparing $e^2$ and $2^e$ is the same as comparing $\left (x^{\frac{1}{x}}\right)^{2e}$ when $x=e$, and $x=2$. Let $f(x)=x^{\frac{1}{x}}=e^{\frac{\ln(x)}{x}}$. To find the greatest value solve $f'(x)=0$, so that $$x^{\frac{1}{x}}\left (\frac{1}{x^2}-\frac{\ln(x)}{x^2}\right)=0$$ which happens when $x=e$. $f''(e)<0$, so this is the maximum value. Hence $$e^2=\left (e^{\frac{1}{e}}\right)^{2e}>\left (2^{\frac{1}{2}}\right)^{2e}=2^e$$
More generally, this shows that $e^x> x^e$, which in particular means that $e^{\pi}>\pi^e$.
$f(x)=\frac{\log x}{x}$ over $\mathbb{R}^+$ has an absolute maximum at $x=e$, since $f'(e)=0$, $e$ is the only root of $f'(x)$ and $\lim_{x\to +\infty}f(x)=0, \lim_{x\to 0^+}f(x)=-\infty$. In particular $f(2)<f(e)$: $$ \frac{\log 2}{2}<\frac{\log e}{e}\Longleftrightarrow e\log 2< 2\log e\Longleftrightarrow 2^e<e^2.$$