Prove that $\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2}\leq\frac{3}{2}$ 
For any reals $a$, $b$, $c$ and $d$ prove that:
$$\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2}\leq\frac{3}{2}$$

C-S in the IMO 2001 stile does not help here:
\begin{align}
&\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2} \\
\leq&\sqrt{4\left(\tfrac{a^2}{(1+a^2)^2}+\tfrac{b^2}{(1+a^2+b^2)^2}+\tfrac{c^2}{(1+a^2+b^2+c^2)^2}+\tfrac{d^2}{(1+a^2+b^2+c^2+d^2)^2}\right)}\\
\leq&2\sqrt{\tfrac{a^2}{1+a^2}+\tfrac{b^2}{(1+a^2)(1+a^2+b^2)}+\tfrac{c^2}{(1+a^2+b^2)(1+a^2+b^2+c^2)}+\tfrac{d^2}{(1+a^2+b^2+c^2)(1+a^2+b^2+c^2+d^2)}} \\
=&2\sqrt{1-\tfrac1{1+a^2}+\tfrac1{1+a^2}-\tfrac1{1+a^2+b^2}+\tfrac1{1+a^2+b^2}-\tfrac1{1+a^2+b^2+c^2}+\tfrac1{1+a^2+b^2+c^2}-\tfrac1{1+a^2+b^2+c^2+d^2}} \\
<&2
\end{align}
We can assume that our variables are non-negative, of course.
For two variables we can get a best estimation here:
$$\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}\leq\sqrt{\frac{207+33\sqrt{33}}{512}}\approx0.88...$$
There is also the following Ji Chen's estimation:
$$\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\dotsb+\frac{x_n}{1+x_1^2+x_2^2+\dotsb+x_n^2}<\sqrt{n}-\dfrac{\ln{n}}{2\sqrt{n}},$$
but it does not help.
Thank you!
 A: A proof for $n=2$.
We need to prove that:
$$\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}\leq\sqrt{\frac{207+33\sqrt{33}}{512}}.$$
Since $x\leq|x|$ and for $ab=0$ it's obvious,  it's enough to prove this inequality for positive variables.
Now, by AM-GM
$$\frac{b}{1+a^2+b^2}=\frac{1}{\frac{1+a^2}{b}+b}\leq\frac{1}{2\sqrt{1+a^2}}$$ and it's enough to prove that $f(a)\leq\sqrt{\frac{207+33\sqrt{33}}{512}},$ where
$$f(a)=\frac{a}{1+a^2}+\frac{1}{2\sqrt{1+a^2}}=\frac{2a+\sqrt{1+a^2}}{2(1+a^2)}.$$
Now, $$f'(a)=\tfrac{\left(2+\frac{a}{\sqrt{1+a^2}}\right)(1+a^2)-\left(2a+\sqrt{1+a^2}\right)\cdot2a}{2(1+a^2)^2}=\tfrac{2-2a^2-a\sqrt{1+a^2}}{2(1+a^2)^2}.$$
Now, $f'(a)=0$ gives $$2(1-a^2)=a\sqrt{1+a^2}$$ and we see that should be $1-a^2>0$.
Thus, $$4(1-2a^2+a^4)=a^2+a^4$$ or
$$3a^4-9a^2+4=0$$ or
$$a^2=\frac{9-\sqrt{33}}{6}$$ or
$$a=\sqrt{\frac{9-\sqrt{33}}{6}}.$$
Now, easy to see that for this value $f$ gets a maximal value and
$$f\left(\sqrt{\tfrac{9-\sqrt{33}}{6}}\right)=\tfrac{2\sqrt{\tfrac{9-\sqrt{33}}{6}}+\sqrt{1+\frac{9-\sqrt{33}}{6}}}{2\left(1+\tfrac{9-\sqrt{33}}{6}\right)}=\tfrac{2\sqrt{9-\sqrt{33}}+\sqrt{15-\sqrt{33}}}{\sqrt2(5\sqrt3-\sqrt{11})}=\sqrt{\tfrac{207+33\sqrt{33}}{512}}.$$
The last equality it's nice.
We can get a proof by the following Carl Schildkraut's beautiful idea:
Why $\sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}=\sqrt{71-17\sqrt{17}}$ is true?
A: First, we give some auxiliary results (Facts 1 through 3). The proofs are easy and thus omitted.
Fact 1: Let $a, b$ be reals. Then $\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}\le \sqrt{\frac{207+33\sqrt{33}}{512}}$.
Fact 2: Let $\gamma$ be real. Then $\frac{\gamma}{1 + \gamma^2} + \sqrt{\frac{207+33\sqrt{33}}{512}}\frac{1}{\sqrt{1+\gamma^2}} < \frac{6}{5}$.
Fact 3: Let $a$ be real. Then $\frac{a}{1+a^2} + \frac{6}{5}\frac{1}{\sqrt{1+a^2}} < \frac{3}{2}$.
$\phantom{2}$
Now, let $\alpha = \frac{c}{\sqrt{a^2 + b^2 + 1}}$ and $\beta = \frac{d}{ \sqrt{a^2 + b^2 + 1}}$.
We have
\begin{align}
&\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2} \\ 
=\ & \frac{\alpha \sqrt{a^2 + b^2 + 1} }{1+a^2+b^2+\alpha^2(a^2 + b^2 + 1)}\\
&\quad + \frac{\beta \sqrt{a^2 + b^2 + 1}}{1+a^2+b^2+\alpha^2(a^2 + b^2 + 1) +\beta^2(a^2 + b^2 + 1)}\\
=\ & \left(\frac{\alpha}{1 + \alpha^2} + \frac{\beta}{1 + \alpha^2 + \beta^2}\right)\frac{1}{\sqrt{a^2+b^2+1}}\\
\le\ & \sqrt{\frac{207+33\sqrt{33}}{512}}\frac{1}{\sqrt{a^2+b^2+1}}
\end{align}
where we have used Fact 1.
Let $\gamma = \frac{b}{\sqrt{1+a^2}}$. We have
\begin{align}
&\frac{a}{1+a^2} + \frac{b}{1+a^2+b^2} + \sqrt{\frac{207+33\sqrt{33}}{512}}\frac{1}{\sqrt{a^2+b^2+1}}\\
=\ & \frac{a}{1+a^2} + \frac{\gamma \sqrt{1+a^2}}{1+a^2+\gamma^2(1+a^2)} + \sqrt{\frac{207+33\sqrt{33}}{512}}\frac{1}{\sqrt{a^2+\gamma^2(1+a^2)+1}}\\
=\ & \frac{a}{1+a^2} + \left(\frac{\gamma}{1 + \gamma^2} + \sqrt{\frac{207+33\sqrt{33}}{512}}\frac{1}{\sqrt{1+\gamma^2}}\right)\frac{1}{\sqrt{1+a^2}}\\
<\ & \frac{a}{1+a^2} + \frac{6}{5}\frac{1}{\sqrt{1+a^2}}\\
<\ & \frac{3}{2}
\end{align}
where we have used Facts 2 and 3.
We are done.
