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.
Best Answer
By the Cauchy–Schwarz inequality \begin{align*} \int_0^1 {(1 + t^2 )^{7/2} {\rm d}t} & \le \sqrt {\int_0^1 {(1 + t^2 )^3 {\rm d}t} \int_0^1 {(1 + t^2 )^4 {\rm d}t} } = \sqrt {\frac{{42496}}{{3675}}} \\ & = \sqrt {\frac{{49}}{4}\frac{{169984}}{{180075}}} < \sqrt {\frac{{49}}{4}} = \frac{7}{2}. \end{align*} Alternatively, $$ \sqrt {\frac{{42496}}{{3675}}} < \sqrt {\frac{{42849}}{{3600}}} = \sqrt {\frac{{4761}}{{400}}} = \frac{{69}}{{20}} < \frac{{70}}{{20}} = \frac{7}{2}. $$