<s>[INST] Proving Definite Integral Strong Inequation I got the following question:
Prove:
$$\int^{1}_0 \frac{x^9}{\sqrt{1+x}} < \frac{1}{10}$$
I proved the weak inequation like so:
$x \geq 0 \Rightarrow \sqrt{1+x} \geq 1$ (monotonicity of square root)
$\Rightarrow \frac{x^9}{\sqrt{1+x}} \leq x^9$
$\Rightarrow \int^{1}_0 \frac{x^9}{\sqrt{1+x}} \leq \int^{1}_0 x^9 = \frac{x^{10}}{10}|_0^1 = \frac{1}{10}$ (integral monotonicity)
However, I am stumped as how to prove a strong inequation here. There doesn't seem to be a number less than $\frac{1}{10}$ that I can limit this function to, so that technique doesn't work, and finding the actual integral is not an option.
 [/INST] If you like something stronger, consider that $\frac{1}{\sqrt{1+x}}$ is a convex function over $[0,1]$, hence:
$$\frac{1}{\sqrt{1+x}}\leq 1-\left(1-\frac{1}{\sqrt{2}}\right)x\tag{1}$$
and:
$$\int_{0}^{1}\frac{x^9}{\sqrt{1+x}}\,dx\leq \int_{0}^{1}x^9\,dx-\left(1-\frac{1}{\sqrt{2}}\right)\int_{0}^{1}x^{10}\,dx = \frac{1}{10}-\frac{1}{11}\left(1-\frac{1}{\sqrt{2}}\right),$$
so:
$$\int_{0}^{1}\frac{x^9}{\sqrt{1+x}}\,dx\leq\frac{1}{11\sqrt{2}}+\frac{1}{110}<\frac{1}{13}.\tag{2}$$
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