In triangle $ABC$, let $x,y,z,k>0$, prove that $$x\sin A+y\sin B+z\sin C\leqslant \frac{\sqrt{\left( x^2+k \right) \left( y^2+k \right) \left( z^2+k \right)}}{k}$$ where $k$ satisfies $\frac{x^2}{x^2+k}+\frac{y^2}{y^2+k}+\frac{z^2}{z^2+k}=1$.
The inequality holds when $\sin A:\sin B:\sin C=\frac{x}{x^2+k}:\frac{y}{y^2+k}:\frac{z}{z^2+k}$.
It is said that it's generalization of $x^2+y^2+z^2\geqslant 2xy\cos C+2yz\cos A+2xz\cos B$.
Thanks in advance.
Suppose $\cos B=t\in(-1,1)$ and we shall use the trick of the expansion of sin C in A, B and organizing it in the form of $A_1 \sin A+ A_2\cos A +A_3$ so that we can use the Cauchy inequality to turn it into a one-variable problem. \begin{aligned} x\sin A+y\sin B+z\sin C&=\left( z\cos B+x \right) \cdot \sin A+z\sin B\cdot \cos A+y\sin B \\ &\le \sqrt{\left( z\cos B+x \right) ^2+z^2\sin ^2B}+y\sin B \\ &=\sqrt{z^2+x^2+2xz\cos B}+y\sin B \\ &=\sqrt{z^2+x^2+2xzt}+y\sqrt{1-t^2} \end{aligned}
(note that the '=' here can actually be obtained due to the fact that $x,y,z,k$ are positive and $\tan A\in \mathbb{R}$). The following work can be done by solving the zeroes in a $3^{rd}$ degree equation (which is the same degree as the condition itself). I tested several combinations, and the result seems to be correct.
Nevertheless, I believe the problem has its own more elegent solutions.