Prove that $\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2}\leq\frac{3}{2}$

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 $\frac{a}{1+a^2} + \frac{6}{5}\frac{1}{\sqrt{1+a^2}} < \frac{3}{2}$.

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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.