How to prove this inequality $xy\sin^2C+yz\sin^2A+zx\sin^2B\le\dfrac{1}{4}$

Question

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.

Answer 1

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 $g(x,y)\leq{1\over4}$.