An inequality using convex functions: If $\frac{a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2} = 1$ then $abc \le \frac1{2\sqrt 2}$ I see this in a Chinese convex analysis book.

Suppose that $a$,$b$,$c$ are positive real numbers satisfying
\begin{equation}
\frac{a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2} = 1.
\end{equation}
Show that $abc \le \dfrac1{2\sqrt 2}$.

Since it's from a convex analysis book, I tried proving this using Jensen's inequality.  However, I can't think of a suitable convex function.  Therefore, I tried AM–GM, but I can't get a product $abc$.
$$(abc)^2\le\frac18\iff8(abc)^2\le1$$
$$\iff\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\right)^3 (abc)^2 \le 1$$
Finally, I used Lagrange multiplier to solve the problem, but I think there is some more elementary solution.
$$f(a,b,c)=abc$$
$$g(a,b,c)=\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}-2=0$$
$$\nabla f(a,b,c)=(bc,ca,ab)$$
$$\nabla g(a,b,c)=\left(-\frac{2a}{(1+a^2)^2},-\frac{2b}{(1+b^2)^2},-\frac{2c}{(1+c^2)^2}\right)$$
$$\because \nabla f = \lambda \nabla g$$
$$\therefore bc = -\frac{2a\lambda}{(1+a^2)^2} \iff abc = -\frac{2a^2\lambda}{(1+a^2)^2}$$
$$abc = -\frac{2a^2\lambda}{(1+a^2)^2} = -\frac{2b^2\lambda}{(1+b^2)^2} = -\frac{2c^2\lambda}{(1+c^2)^2}$$
$$\frac{a}{1+a^2}=\frac{b}{1+b^2}=\frac{c}{1+c^2}=\mu$$
$$a+\frac1a=b+\frac1b=c+\frac1c$$
$$\because \frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}=2$$
$$\therefore \frac{\mu}{a}+\frac{\mu}{b}+\frac{\mu}{c}=2$$
$$\because \frac{a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2} = 1$$
$$\therefore a\mu + b\mu + c\mu = 1$$
$$\frac1a+\frac1b+\frac1c=2(a+b+c)$$
$$3(a+b+c)=a+\frac1a+b+\frac1b+c+\frac1c=3\left(a+\frac1a\right)$$
$$b+c=\frac1a$$
Similarly, $c+a=\dfrac1b$ and $a+b=\dfrac1c$.  Substitute $a=\dfrac1{b+c}$ into the other two equations.
$$c+\frac1{b+c}=\frac1b$$
$$\frac1{b+c}+b=\frac1c$$
$$b(b+c)c+b=b+c$$
$$c+b(b+c)c=b+c$$
Subtracting one equation from another, we get $b=c$.  Similarly, we have $a=b=c$.  It remains to substitute it back to the original constraint and calculate the product.
Any alternative solution is appreciated.
 A: One possibility is to denote $A=\frac{a^2}{1+a^2}$, etc., so $a^2=\frac{A}{1-A}$, etc. We are given $A+B+C=1$ and want to prove that $8ABC\le(1-A)(1-B)(1-C)$, or, after opening the parentheses, replacing the sum by $1$, and putting all $ABC$ on the left,
$$
9ABC\le AB+BC+AC=(AB+BC+AC)(A+B+C)\,.
$$
Now just use AM-GM for each expression in parentheses on the right separately.
A: I believe I can get you started at least.  Break down your equation a little.
$a,b,c \in \mathbb{R}^+ \rightarrow \frac{1}{1+n^2} \in [0,1],\;\;\; n = a,b,c$ 
This means that your original equation could be considered a convex combination as follows: $a^2\lambda_1 + b^2\lambda_2 +c^2\lambda_3 = 1$, where the $\lambda$ values are the fractions defined above.   
That, however, is just part of the setup.  Basically you've determined that $\frac{1}{1+a^2} + \frac{1}{1+b^2} + \frac{1}{1+c^2} = 1$, by definition of convex combination.  My interpretation is that what you're really being asked to prove is what is the minimum value that each $\lambda_i$ can take in a convex combination.  This is because, given the definition of $\lambda_i$ for this problem, when a $\lambda$ reaches a minimum, it is because the variable ($a,b,c$ for a given $\lambda$) will be at a maximum.  When you maximize $a,b,$ and $c$ then you also maximize their product $abc$.
Consider the case with just two variables between 0 and 1 that must also sum to 1.  You want to minimize them both, so the only possible solution is that they equal each other.  Neither can be improved without departing from the constraint of the problem (namely that they sum to 1).  This extends quite easily to 3 variables, and there are plenty of proofs to look up online.
Back to the specifics of your problem: Since all of the coefficient expressions are equivalent, you can assume $a = b = c$ at the extreme value.  Knowing this:
$\frac{3a^2}{1+a^2} = 1 \rightarrow a = \frac{1}{\sqrt{2}}$ and $abc = (\frac{1}{\sqrt{2}})^3 = \frac{1}{2\sqrt{2}} $
A: By C-S $$1=\sum_{cyc}\frac{a^2}{1+a^2}\geq\frac{(a+b+c)^2}{3+a^2+b^2+c^2},$$
which by AM-GM gives
$$3\geq2(ab+ac+bc)\geq6\sqrt[3]{a^2b^2c^2}$$
and we are done!
