How to prove that $3^\pi > \pi^3$

Question

I need to prove this inequality that $3^\pi > \pi^3$ How can i start to answer this problem. What concept should I apply?

Answer 1

One way is to prove that the function $$ x\mapsto x^{1/x} $$ is monotonically decreasing if $x>e$. I leave the details to you.

Answer 2

I would write $\pi = 3 +\epsilon$, where $\epsilon \gt 0$. Then rewrite as follows:

$$\begin{align}3^{\pi} &= 3^{3+\epsilon} = 3^3 \cdot 3^{\epsilon} = 3^3 \cdot e^{\epsilon \log{3}}\\ &\ge 3^3 \left (1+(\log{3}) \epsilon + \frac12 (\log{3})^2 \epsilon^2 +\frac16 (\log{3})^3 \epsilon^3\right ) \end{align}$$

$$\begin{align}\pi^3 = (3+\epsilon)^3 &= 3^3 + 3 \cdot 3^2 \epsilon + 3 \cdot 3 \epsilon^2+\epsilon^3\\ &= 3^3 \left (1+\epsilon+\frac13 e^2 + \frac1{27} e^3 \right )\end{align}$$

What can you conclude from this? NB $\log{3} \gt 1$.

You should also prove that $\log{3} \gt 1$ by showing that

$$\left ( 1+\frac1{n} \right )^n \lt 3 $$

for all $n$.

Answer 3

Take logarithms and use the fact that $f(x)=\frac{\ln x}{x}$ is strictly decreasing for $x>e$ while $\ln x, e^x$ are strictly increasing:

$$3 < \pi$$ $$\Longrightarrow \frac{\ln 3}{3} > \frac{\ln \pi}{\pi}$$ $$\Longrightarrow \pi \ln3 > 3 \ln \pi$$ $$\Longrightarrow 3^\pi > \pi^3$$

Answer 4

Write $\pi = 3+\epsilon$, then by binomial theorem $$\pi^3 = 3^3+3\cdot3^2\epsilon + 3\cdot3^1\epsilon^2 + \epsilon^3$$ we can rewrite this as $$3^3(1+\epsilon+\epsilon^2/3 + \epsilon^3/3^2)$$

Then consider $3^\pi = 3^{(3+\epsilon)} = 3^3 3^\epsilon$

Now the taylor expansion for $3^\epsilon$ is $1 + \log(3)\epsilon + \log(3)^2\epsilon^2/2 + \log(3)^3\epsilon^3/6 + \cdots$

Pairwise comparison of terms, that all terms in exponential are non-negative and $\log(3)>1$ should show that the exponential is larger.