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?

By the law of sines and AM-GM we obtain $$5c^2=a^2+b^2\geq2ab,$$ which gives
$$\frac{c^2}{ab}\geq\frac{2}{5}.$$
In another hand, by the law of cosines we obtain:
$$\cos{C}=\frac{a^2+b^2-c^2}{2ab}=\frac{2c^2}{ab}\geq\frac{4}{5}.$$
Id est, $$\sin{C}=\sqrt{1-\cos^2C}\leq\sqrt{1-\left(\frac{4}{5}\right)^2}=\frac{3}{5}.$$
