How to prove: $2^\frac{3}{2}<\pi$ without writing the explicit values of $\sqrt{2}$ and $\pi$ How to prove: $2^\frac{3}{2}<\pi$  without writing the explicit values of $\sqrt{2}$ and $\pi$.
I am trying by calculus but don't know how to use here in this problem. Any idea?
 A: Let $ABCD$ be a square inscribed in the unit circle. Then the length of arc $AB$ is $\cfrac{\pi}{2}$, and the length of side $AB$ is $\sqrt{2}$. Since the shortest path between two points is the straight line segment between them, it follows that $\sqrt{2} \lt \cfrac{\pi}{2} \;\iff\; \sqrt{8} \lt \pi\,$.
A: It's pretty easy to see that $\pi > 3$, by inscribing a circle inside a regular hexagon. Then squaring both sides gives $\pi^2>9>8$. Taking square roots again gives $\pi > \sqrt{8}=2^\frac{3}{2}$
A: Say $\sin x\leq x$ with $x=\dfrac{\pi}{6}$ we get $\dfrac12\leq\dfrac{\pi}{6}$ or $3\leq\pi$ and $8<9\leq\pi^2$.
A: If we take the definition of $\pi$ to be based on the circumference of a circle (namely that $\frac{C}{d} = \pi$), then we can see a nice geometric proof of this fact.
Consider a square with side length $1$. Now, draw a circle around this square such that all four corners of the square fall on the circle. This circle has a diameter of $\sqrt{2}$. The perimeter of the square involved is $4$.  
$\frac{4}{\sqrt{2}} = 2^{\frac{3}{2}}$
Now, by looking at this picture it is quite obvious that the circumference of the circle is larger than the perimeter of the square. If you were constructing these shapes this could be easily verified. But now we can say that:
\begin{align}
C &> P_{square} \\
C &> 4 \\
\frac{C}{d} &> \frac{4}{d} \ \ \ \ \textbf{Since $d > 0$} \\
\pi &> 2^{\frac{3}{2}}
\end{align}
A: Notice that $\sin x<x$ on $(0,\pi/2)$ by applying the Mean Value Theorem on $\sin x - \sin 0$ (and noting that $\cos t<1$ whenever $0<t<x<\pi/2$). Hence
$$ \int\limits_{0}^{\pi/2}{\sin x\,dx} < \int\limits_{0}^{\pi/2}{x\,dx}. $$
Notice that the left-hand side is $1$, while the right-hand side is $\pi^2/8$.
