How to prove this inequality $xy\sin^2C+yz\sin^2A+zx\sin^2B\le\dfrac{1}{4}$ Let $x,y,z$ is real numbers,and such that $x+y+z=1$,and in $\Delta ABC$,prove that
$$xy\sin^2C+yz\sin^2A+zx\sin^2B\le\dfrac{1}{4}$$
I think this inequality maybe use $x^2+y^2+z^2\ge 2yz\cos{A}+2xz\cos{B}+2xy\cos{C},x,y,z\in R$
Thank you everyone.
 A: I take $x$ and $y$ as independent variables and put $z:=1-x-y$. 
For the moment, let $a$, $b$, $c$ be arbitrary positive constants, and consider the function
$$g(x,y)= c x y + (a y + b x)(1-x-y)\ .$$
It has a single critical point $$(x_0,y_0)=\left({a(b+c-a)\over \Delta},\ {b(a+c-b)\over\Delta}\right)\ ,\quad\Delta:=2(a b+b c+c a)-(a^2+b^2+c^2)\ .$$
We therefore look at the function
$$q(u,v):=g(x_0,y_0)-g(x_0+u,y_0+v)=b u^2+(a+b-c)uv+a v^2\ .$$
We have $a>0$ and $\det(q)= ab-{1\over4}(a+b-c)^2={1\over4}\Delta$. Therefore, when $\Delta>0$, the quadratic form $q$ is positive definite:  $q(u,v)>0$ for all $(u,v)\ne(0,0)$, which implies that $g$ assumes a global maximum at $(x_0,y_0)$. It follows that when $\Delta>0$ we are sure that
$$g(x,y)\leq g(x_0,y_0)={a b c\over\Delta}\ .$$
Now in our case
$$a=\sin^2\alpha\ ,\quad b=\sin^2\beta,\quad  c=\sin^2\gamma=\sin^2(\alpha+\beta)\ .$$
A simple calculation shows that we indeed have
$$\Delta=4\sin^2\alpha\sin^2\beta\sin^2(\alpha+\beta)=4 a b c>0\ ,$$
so that we arrive at the stated inequality $g(x,y)\leq{1\over4}$.
A: We need to prove that $$(x+y+z)^2\geq4xy\sin^2\gamma+4xz\sin^2\beta+4yz\sin^2\alpha$$ or
$$x^2+y^2+z^2+2xy\cos2\gamma+2xz\cos2\beta+2yz\cos2\alpha\geq0$$ or
$$z^2+2(x\cos2\beta+y\cos2\alpha)z+x^2+y^2+2xy\cos2\gamma\geq0,$$
for which it's enough to prove that
$$(x\cos2\beta+y\cos2\alpha)^2-(x^2+y^2+2xy\cos2\gamma)\leq0$$ or
$$x^2\sin^22\beta+2(\cos2\gamma-\cos2\beta\cos2\alpha)xy+y^2\sin^22\alpha\geq0$$ or
$$(x\sin2\beta-y\sin2\alpha)^2\geq0.$$
Done!
