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.
 A: 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,
$$
\sqrt {\frac{{42496}}{{3675}}}  < \sqrt {\frac{{42849}}{{3600}}}  = \sqrt {\frac{{4761}}{{400}}}  = \frac{{69}}{{20}} < \frac{{70}}{{20}} = \frac{7}{2}.
$$
A: 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.
A: Since $t^2 \in [0,1],\ $ we may use the Binomial expansion,
$$ \left( 1+ t^2 \right)^{7/2} = 1 + \frac{7}{2}t^2 + \frac{35}{8} t^4 + \frac{35}{16} t^6 + \frac{35}{128} t^8 + (\text{ alternating sequence of decreasing terms with negative leading term }),$$
so,
$$ \left( 1+ t^2 \right)^{7/2} < 1 + \frac{7}{2}t^2 + \frac{35}{8} t^4 + \frac{35}{16} t^6 + \frac{35}{128} t^8\qquad \forall\ t\in(0,1) $$
Unless I've made a calculation error, I get:
$$\ \int_0^1 1 + \frac{7}{2}t^2 + \frac{35}{8} t^4 + \frac{35}{16} t^6 + \frac{35}{128} t^8\ dt < 3.4.$$
