Not Your Ordinary System of Inequalities

$3(x-1)^2-x^2$

$=3(x^2-2x+1)-x^2$

$=2x^2-6x+3$

$x(x-2)+(x-1)(x-3)$

$=x^2-2x+x^2-4x+3$

$=2x^2-6x+3$

$\therefore 3(x-1)^2-x^2=x(x-2)+(x-1)(x-3)$

$\because x(x-2)\geq0$ and $(x-1)(x-3)\geq0$

$\therefore x(x-2)+(x-1)(x-3)\geq0$

$3(x-1)^2-x^2\geq0$

$\cfrac{576}{x(x-2)}+\cfrac{225}{(x-1)(x-3)} \leq \cfrac{1521}{3(x-1)^2-x^2}$

$\Rightarrow [3(x-1)^2-x^2][\cfrac{576}{x(x-2)}+\cfrac{225}{(x-1)(x-3)}] \leq1521$

$\Rightarrow [x(x-2)+(x-1)(x-3)][\cfrac{576}{x(x-2)}+\cfrac{225}{(x-1)(x-3)}] \leq1521$

By Cauchy-Schwarz Inequality ,

$\Rightarrow [x(x-2)+(x-1)(x-3)][\cfrac{576}{x(x-2)}+\cfrac{225}{(x-1)(x-3)}] \geq1521$

$\therefore$ we conclude that there exists solution if and only if $[x(x-2)+(x-1)(x-3)][\cfrac{576}{x(x-2)}+\cfrac{225}{(x-1)(x-3)}]=1521$

Also, the equality holds if and only if

$x(x-2)=\cfrac{576}{x(x-2)}$ $\textbf{and}$ $(x-1)(x-3)=\cfrac{225}{(x-1)(x-3)}$

$[x(x-2)]^2=576$

$[x(x-2)]=\pm 24$

$\because x(x-2)\geq0$

$\therefore x(x-2)=24$

$\Rightarrow x^2-2x-24=0$

$(x-6)(x+4)=0$

$x=6$ or $x=-4$

$[(x-1)(x-3)]^2=225$

$(x-1)(x-3)=\pm15$

$\because(x-1)(x-3)\geq0$

$\therefore(x-1)(x-3)=15$

$\Rightarrow x^2-4x-12=0$

$(x-6)(x+2)=0$

$x=6$ or $x=-2$

We conclude that the solution, which is common to the two equations solved, is $\boxed{6}$ .