Prove: For $\triangle ABC$, if $\sin^2A + \sin^2B = 5\sin^2C$, then $\sin C \leq \frac{3}{5}$. We have a triangle $ABC$. It is given that $\sin^2A + \sin^2B = 5\sin^2C$. Prove that $\sin C \leq \frac{3}{5}$.
Let's say that $BC = a$, $AC=b$, $AB=c$.
According to the sine law,
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,$$ then
$\sin A = \frac{a}{2R}\implies \sin^2A = \frac{a^2}{4R^2}$
$\sin B = \frac{b}{2R}\implies \sin^2B = \frac{b^2}{4R^2}$
$\sin C = \frac{c}{2R}\implies \sin^2C = \frac{c^2}{4R^2}$
Then we get:
$$\frac{a^2}{4R^2} + \frac{b^2}{4R^2} = 5\frac{c^2}{4R^2}.$$
Since $4R^2 > 0$, we get that $a^2 + b^2 = 5c^2$
Guys, is that correct? Even if it is, do you have any ideas what shall I do next?

Since $A + B + C = \pi$, we have $\sin (A + B) = \sin C$.
Also notice that
$$\cos^2 A + \cos^2 B = 1 - \sin^2A + 1 - \sin^2B = 2 - 5\sin^2(A + B)$$
Proceeding by CSB, we get:
\begin{align}
\sin(A + B) &= \sin A \cos B + \sin B \cos A \\
&\le \sqrt{\sin^2 A + \sin ^2 B} \sqrt{\cos^2A + \cos^2B} \\
&= \sqrt{5} \sin(A + B) \sqrt{2 - 5\sin^2(A + B)}
\end{align}
Therefore $$1 \le 5(2 - 5\sin^2(A + B)) = 10 - 25\sin^2(A + B)$$ which yields $$\sin C = \sin(A + B) \le \frac35$$
