The maximum for $xy \sin \alpha + yz \sin \beta +zx \sin \gamma$. Question: 
Deduce the maximum of $xy \sin \alpha + yz \sin \beta +zx \sin \gamma$ if $x,y,z$ are real numbers that satisfy $x^2+3y^2+4z^2=6$ with $0<\alpha,\beta,\gamma<\pi$ such that $\alpha+\beta+\gamma=2\pi$.
Currently, I am not very sure how to approach the problem. I had an idea to consider the area of a triangle made of 3 smaller triangles with areas $\frac{1}{2}xy \sin \alpha$, $\frac{1}{2}yz \sin \beta$ and $\frac{1}{2}zx \sin \gamma$ respectively. However, that kinda got me no where as I did not have any good ideas on how to use the condition $x^2+3y^2+4z^2=6$. Moreover, that would have assumed $x,y,z\geq0$ which might not be the case.
So, is there a way to deduce the maximum without a calculus approach?
 A: Okay I finally arrived at an answer. First, lets reduce the cases we need to look at. Clearly $\sin \alpha, \sin \beta, \sin \gamma >0$. So, if one/two of the variables $x,y$ or $z$ is negative, the expression $xy \sin \alpha +yz \sin \beta + zx \sin \gamma$ will not be maximum. Clearly, if all three $x,y,z$ are negative, it is the same when they are positive for the expression $xy \sin \alpha +yz \sin \beta + zx \sin \gamma$. This means we can reduce to the problem to only when $x,y,z>0$.
Consider the following triangle;

Let $CD=x, BD=y, AD=z$ with $\angle CDB=\alpha,\angle BDA=\beta$ and $\angle ADC=\gamma$. This implies that the area of $ABC$, I'll denote it as $[ABC]$, will be $\frac{1}{2}(xy \sin \alpha +yz \sin \beta + zx \sin \gamma)$. 
Let $DE,CF$ be altitudes with respect to $AB$, where $E,F$ are on $AB$. Let $AE=m, BE=n, DE=h$ with $CF=H$. This implies that $z^2=m^2+h^2$ and $y^2=n^2+h^2$.
Okay now we are gonna bash this with inequalities,
$$\begin{align} 
6 &=x^2+3y^2+4z^2\\ 
&=x^2+3h^2+3n^2+4h^2+4m^2\\
&=x^2+7h^2+3n^2+4m^2\\
& \geq \frac{(x+h)^2}{1+\frac{1}{7}}+\frac{(n+m)^2}{\frac{1}{3}+\frac{1}{4}}\tag{1}\label{eq1} \\
& \geq 2\sqrt{\frac{(x+h)^2(n+m)^2}{\frac{8}{7}\times\frac{7}{12}}} \tag{2}\label{eq2} \\
& =2\sqrt{\frac{3}{2}}(x+h)(m+n)\\
&\geq \sqrt{6} (H)(m+n)\tag{3}\label{eq3} \\
&=2\sqrt{6} [ABC].
\end{align}$$
Now $\eqref{eq1}$ is due to Cauchy Schwarz Inequality, $\eqref{eq2}$ is due to AM-GM Inequality and $\eqref{eq3}$ is due to Triangle Inequality. I guess equality does occur for each case.
So this implies that;
$$\begin{align}
& 6\geq 2\sqrt{6} [ABC]\\
\Longleftrightarrow & \sqrt{6} \geq 2[ABC]=xy \sin \alpha +yz \sin \beta + zx \sin \gamma.
\end{align}$$
Therefore, the maximum of $xy \sin \alpha +yz \sin \beta + zx \sin \gamma=\sqrt{6}$.
