<s>[INST] How to show that $\int_0^1 (1+t^2)^{\frac 7 2} dt < \frac 7 2 $? I need to show that $\int_0^1 (1+t^2)^{\frac 7 2} dt < \frac 7 2 $. I've checked numerically that this is true, but I haven't been able to prove it.
I've tried trigonometric substitutions. Let $\tan u= t:$
$$\int_0^1 (1+t^2)^{\frac 7 2} dt = \int_0^{\frac{\sqrt 2}{2}} (1+\tan^2 u )^{\frac 9 2} du = \int_0^{\frac{\sqrt 2}{2}} \sec^9 u \ du = \int_0^{\frac{\sqrt 2}{2}} \sec^{10} u \cos u\ du = \int_0^{\frac{\sqrt 2}{2}} \frac {\cos u}{(1-\sin^2 u)^5} du$$
Now let $\sin u = w$. Then:
$$\int_0^{\frac{\sqrt 2}{2}} \frac {\cos u}{(1-\sin^2 u)^5} du = \int_0^{\sin {\frac{\sqrt 2}{2}}} \frac {1}{(1-w^2)^5} dw.$$
This last integral is solvable using partial fraction decomposition, but even after going through all the work required I'm not really sure how to compare it with $\frac 7 2$, because of that $\sin {\frac {\sqrt{2}}{2}}$ term, which is not easy to compare.
 [/INST] Alternative proof:
Clearly, we have
$\sqrt{1 + t^2} \le 1 + t^2/2$. Thus, we have
$$(1 + t^2)^{7/2} \le (1 + t^2)^3 (1 + t^2/2) = \frac12t^8 + \frac52 t^6 + \frac92 t^4 + \frac72 t^2 + 1.$$
Thus, we have
\begin{align*}
 \int_0^1 (1 + t^2)^{7/2} \, \mathrm{d} t
 &\le \int_0^1 \left( \frac12t^8 + \frac52 t^6 + \frac92 t^4 + \frac72 t^2 + 1\right)\,\mathrm{d} t \\
&= \frac{1}{18} + \frac{5}{14}
 + \frac{9}{10} + \frac{7}{6} + 1\\
 & < \frac72.
\end{align*}
We are done.
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