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.

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.
