4. Given that $a, b, c$ are the lengths of the three sides of $\triangle ABC$, and satisfy the conditions
$$
\frac{2 a^{2}}{1+a^{2}}=b, \frac{2 b^{2}}{1+b^{2}}=c, \frac{2 c^{2}}{1+c^{2}}=a \text {. }
$$

Then the area of $\triangle ABC$ is . $\qquad$
4. $\frac{\sqrt{3}}{4}$.

From the problem, we have
$$
\begin{array}{l}
\frac{2 a^{2}}{1+a^{2}}=b \Rightarrow 1+\frac{1}{a^{2}}=\frac{2}{b}, \\
\frac{2 b^{2}}{1+b^{2}}=c \Rightarrow 1+\frac{1}{b^{2}}=\frac{2}{c}, \\
\frac{2 c^{2}}{1+c^{2}}=a \Rightarrow 1+\frac{1}{c^{2}}=\frac{2}{a} .
\end{array}
$$
(1) + (2) + (3) gives
$$
\left(1-\frac{1}{a}\right)^{2}+\left(1-\frac{1}{b}\right)^{2}+\left(1-\frac{1}{c}\right)^{2}=0 \text {. }
$$

Thus, $1-\frac{1}{a}=1-\frac{1}{b}=1-\frac{1}{c}=0$.
Therefore, $a=b=c=1$.
Hence, $S_{\triangle A B C}=\frac{\sqrt{3}}{4}$.