How to prove this inequality $x\sin^2{A}+y\sin^2{B}\ge xy\sin^2{C}$

Question

In $\Delta ABC$,if $x,y>0$ and $x+y=1$.show that $$x\sin^2{A}+y\sin^2{B}\ge xy\sin^2{C}$$

I have looked at the simpler methods,? Here is one solution

$$\dfrac{\sin^2{A}}{y}+\dfrac{\sin^2{B}}{x}\ge\dfrac{(\sin{A}+\sin{B})^2}{x+y}=\dfrac{\sin^2{(A+B)}}{1}=\sin^2{C}$$ where use this $\sin{A}+\sin{B}\ge \sin{(A+B)}$

other methods?

Answer 1

It's $\frac{4S^2x}{b^2c^2}+\frac{4S^2y}{a^2c^2}\geq\frac{4S^2xy}{a^2b^2}$ or

$(x+y)(a^2x+b^2y)\geq xyc^2$ or $a^2x^2+b^2y^2+(a^2+b^2-c^2)xy\geq0$.

$\Delta=(a^2+b^2-c^2)^2-4a^2b^2=-16S^2<0$ and we are done!