Maximizing Some Sum

I think you love CS inequality too much ;-P.... Isn't it @Harsh Shrivastava ?

Equation can be rewritten as: $6(xy+yz+zx)=3(x+y+z)^2-65(x+y+z)$

$\small{(\color{#D61F06}{\text{Using } \left(\sum_{cyc} x\right)^2=\sum_{cyc}x^2+2\sum_{cyc}xy })}$

We get-

$3(x^2+y^2+z^2)=65(x+y+z)\color{#20A900}{.....(1)}$

Now applying Cauchy-Schwarz-Inequality :

$\large{\begin{aligned}(x+y+z)^2&\leq 3(x^2+y^2+z^2)\\&=65(x+y+z)~~~\color{#20A900}{\small{\text{Using } (1)}}\end{aligned}}$

$\implies (x+y+z)((x+y+z)-65)\leq 0$

$\Large \boxed{x+y+z\leq 65}$

Since $x+y+z>0$ .

---

Equality occurs when- $\textbf{x=y=z}=\dfrac{65}{3}$

$\frac{xy+yz+zx}{x+y+z}=\frac{3(x+y+z)-65}{6}\Rightarrow \frac{(x+y+z)^2-(x^2+y^2+z^2)}{2(x+y+z)}=\frac{3(x+y+z)-65}{6}\Rightarrow \frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=(x+y+z)\frac{3(x+y+z)-65}{6}$ $\Rightarrow (x+y+z)^2-(x^2+y^2+z^2)=(x+y+z)^2-\frac{65}{3}(x+y+z)\Rightarrow \frac{65}{3}(x+y+z)=x^2+y^2+z^2$ . Then by the QM-AM inequality, we have $x+y+z\leq\sqrt{3(x^2+y^2+z^2)}=\sqrt{65(x+y+z)} \Rightarrow (x+y+z)^2\leq 65(x+y+z)$ . Thus, $x+y+z\leq 65$ and equality is achieved when $x=y=z=\frac{65}{3}$ .