<s>[INST] How to prove this inequality $x\sin^2{A}+y\sin^2{B}\ge xy\sin^2{C}$ 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?
 [/INST] 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!
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