Is my determination of this maximum correct? 

Consider $\Omega:=B_1(0)\subset\mathbb{R}^n$ (it is the open unit ball), $\mathbb{R}^n$ is provided with the euclidean norm $\lVert\cdot\rVert_2$.
    Now I want to determine the following maximum:
    $$
\max\left\{\max_{x\in\partial\Omega}\left\{\lvert \sin^3(x_1)\rvert\right\},\sup_{x\in\Omega}\left\{\frac{\lvert\sum_{i=1}^{n}x_i^2\rvert}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert}\right\}\right\}
$$


Here is my result:
$$
\max_{x\in\partial\Omega}\left\{\lvert \sin^3(x_1)\rvert\right\}=\sin^3(1)\approx0,596
$$
Now to the supremum
$$
\sup_{x\in\Omega}\left\{\frac{\lvert\sum_{i=1}^{n}x_i^2\rvert}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert}\right\}:
$$
I estimated as follows:
$$
\frac{\lvert\sum_{i=1}^{n}x_i^2\rvert}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert}<\frac{\sum_{i=1}^{n}\lvert x_i\rvert^2}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert}\\<\frac{n}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert}
$$
Because of 
$$
\sum_{i=1}^{n}\frac{1}{1+x_i^2}>\frac{n}{2}
$$
on $B_1(0)$, it is
$$
\left\lvert \frac{n}{2}-1-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\right\rvert >1
$$
and therefore 
$$
\frac{n}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert}<n,
$$
so the supremum is $n$.
Because of $n\geq 1$ the searched maximum is $n$.

Please tell me if I am right! Thank you very much!
Sincerely yours,
math12
 A: Notice that: $$\sup\limits_{x\in\Omega}\;\dfrac{\lvert\sum_{i=1}^{n}x_i^2\rvert}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert}=\sup\limits_{x\in\Omega}\;\left\{\dfrac{\sum_{i=1}^{n}x_i^2}{ (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}},-\dfrac{\sum_{i=1}^{n}x_i^2}{ (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}}\right\}$$
Moreover, fix $j$ and let $f_j(x_j)=\dfrac{\sum_{i=1}^{n}x_i^2}{ (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}}$ and so the first-order condition for an interior maximum is: $$f_j^\prime(x_j)=\dfrac{2x_j\cdot\left( (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\right)-\left(\sum_{i=1}^n x_i^2\right)\cdot\left(\dfrac{2x_j}{\left(1+x_j^2\right)^2}\right)}{\left( (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\right)^2}=0\quad \forall j\tag{FOC}$$
The "symmetry" of the problem suggests (or you can prove if $x_i>0$ and $x_j>0$) that $x_i=x_j$ for all $i$ and $j$ at the solution. So the FOC becomes: $$2x\left(\left(\frac{n}{2}-1\right)-\frac{n}{1+x^2}\right)-\dfrac{2x^3}{(1+x^2)^2}=0\Rightarrow x^2=\dfrac{4+\sqrt{n^2+12}}{n-2}.$$ 
Maybe you can solve from here? Here is a list of what is left to do:


*

*Show that $n\cdot \dfrac{4+\sqrt{n^2+12}}{n-2}\le 1$ so our candidate to a solution is on the unit ball.

*Solve the case $n=2$ aside. The above works only for $n\ge 3$.

*If a point in the boundary say $z=(z_i)$ (I ignored the boundary so far) is optimal then $$f_i^\prime(z_i)=f_j^\prime(z_j)\ge 0\text{ for all  } i \text { and } j \text{ such that } z_i>0 \text { and } z_j>0 \text{ and } f_k^\prime(z_k)\le 0 \text{ if } z_k=0.$$

*If there is a point in the boundary that is optimal then I think none of its coordinates can be zero. But in this case all the $x_i^2$ will also be equal, you just have to solve $f_i^\prime(z_i)=\lambda>0$ for all $i$. But before one must prove that no point can be optimal if the square of the coordinates is not the same.

*Notice (for my work above) that the FOC for $f_j(x_k)$ and the FOC for $-f_k(x_k)$ are the same. So I don't have to consider two cases. Only after you compute the $x_i^2$ do you have to figure out the sign of $\left(\frac{n}{2}-1\right)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}$.

