Here is a problem with a cute solution.
Show that the arc length of the parabola $y = x^2$, from $(0,0)$ to $(1,1)$ is not greater than $1.5$.
Scroll down for a solution.
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The arc length of $f(x)$ is given by the integral $\int \sqrt{1 + (f'(x))^2}$.
In our case, we are looking at
$$\int_{0}^{1} \sqrt{1 + 4x^2}$$
This can actually be evaluated without much trouble, and comes out to $\frac{1}{4}(\sqrt{5} + \sinh^{-1}(2)) \approx 1.47$
That is one way of trying to prove the $1.5$ upper bound but we will go for the "cute" proof with very little computations here.
Write
$$\sqrt{1+4x^2} = \sqrt{(2x+1)^2 - 4x} = \sqrt{2x + 1 + 2\sqrt{x}} \sqrt{2x + 1 - 2\sqrt{x}}$$
Let $f(x) = \sqrt{2x + 1 + 2\sqrt{x}}$ and $g(x) = \sqrt{2x + 1 - 2\sqrt{x}}$
We apply the integral version of Cauchy-Schwartz inequality
$$ \int fg \le \sqrt{\int f^2} \sqrt{\int g^2}$$
To get
$$\int_{0}^{1} \sqrt{1+4x^2} \le \sqrt{\int_{0}^{1} (2x + 1 + 2\sqrt{x})} \sqrt{\int_{0}^{1} (2x + 1 -2\sqrt{x})}$$
$$ = \sqrt{1 + 1 + \frac{4}{3}}\sqrt{1 + 1 - \frac{4}{3}} = \sqrt{\frac{20}{9}} < \frac{3}{2}$$
