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.
 A: $\sqrt{1+x}>1$ for every $x > 0$ so $$
\int_0^1\frac{x^9}{\sqrt{1+x}}dx = \int_0^{1/2}\frac{x^9}{\sqrt{1+x}}dx+\int_{1/2}^1\frac{x^9}{\sqrt{1+x}}dx\\
\leq \int_0^{1/2}x^9dx+\int_{1/2}^1\frac{x^9}{\sqrt{1+x}}dx \\ < \int_0^{1/2}x^9dx+\int_{1/2}^1x^9dx =  \int_0^{1}x^9dx$$
A: If $f(x)<g(x)$ for EVERY value of $x$ in $[a,b]$ except one or both of the endpoints, and $a<b$, then
$$
\int_a^b f(x)\,dx <\int_a^b g(x)\,dx.
$$
The proof of this is simple when $f$ and $g$ are continuous, and in fact $x\mapsto\dfrac{x^9}{\sqrt{1+x}}$ and $x\mapsto x^9$ are continuous on $[0,1]$.  One proof is this: If $f(c)<g(c)$ then let $\varepsilon>0$ be less than the distance between $f(c)$ and $g(c)$.  By continuity, there exists $\delta>0$ such that the distance between $f(x)$ and $g(x)$ is less than $\varepsilon$ whenever $x$ is between $a\pm\delta$.  Observe then that
$$
\int_{c-\delta}^{c+\delta} g(x)-f(x)\,dx \ge \int_{c-\delta}^{c+\delta} \varepsilon\,dx=2\delta\varepsilon>0.
$$
A: 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:
$$\int_{0}^{1}\frac{x^9}{\sqrt{1+x}}\,dx\leq\frac{1}{11\sqrt{2}}+\frac{1}{110}<\frac{1}{13}.\tag{2}$$
