Finding a better bound in an inequality Consider points $(x,y)$ on the curve $\sqrt{x^2-3x}+\sqrt{y^2-3y}=1$.
Prove that for all such pairs:
$$x^2+y^2\lt2(x+y)+8.$$

NOTE.- This problem was proposed by two mathematicians, from Romania and Spain, to a math blog in Madrid with the number $15$ on the $RHS$.  In my solution I lowered this number to $8$.
By the way the number $8$ can also be improved, it is not the best bound.   The task at hand is to find the tightest bound.
 A: We'll prove that $$\max_{\sqrt{x^2-3x}+\sqrt{y^2-3y}=1}(x^2+y^2-2(x+y))=\frac{11+\sqrt{13}}{2}.$$
Indeed, let $y\leq0.$ 
Thus, $$x^2-3x\leq1,$$ which gives $$x\leq\frac{3+\sqrt{13}}{2}$$ and
$$x^2+y^2-2(x+y)=x^2-3x+y^2-3y+x+y\leq$$
$$\leq\left(\sqrt{x^2-3x}+\sqrt{y^2-3y}\right)^2+x\leq1+\frac{3+\sqrt{13}}{2}<\frac{11+\sqrt{13}}{2}.$$
Id est, it's enough to prove our inequality for $x\geq3$ and $y\geq3.$
Now, let $\sqrt{x^2-3x}=a$ and $\sqrt{y^2-3y}=b$.
Thus, $a+b=1$, $x=\frac{3+\sqrt{9+4a^2}}{2}$, $y=\frac{3+\sqrt{9+4b^2}}{2}$ and we need to prove that
$$\left(\tfrac{3+\sqrt{9+4a^2}}{2}\right)^2+\left(\tfrac{3+\sqrt{9+4b^2}}{2}\right)^2-2\left(\tfrac{3+\sqrt{9+4a^2}}{2}+\tfrac{3+\sqrt{9+4b^2}}{2}\right)\leq\tfrac{11+\sqrt{13}}{2}$$ or
$$2(a^2+b^2)+\sqrt{9+4a^2}+\sqrt{9+4b^2}\leq5+\sqrt{13}.$$
Now, let $f(x)=\sqrt{9+4x^2}.$
Thus, $$f''(x)=\frac{36}{\sqrt{(9+4x^2)^3}}>0,$$
which says that $f$ is a convex function.
Thus, since for $a\geq b$ we have  $$(a+b,0)\succ(a,b),$$ by Karamata $$f(a)+f(b)\leq f(a+b)+f(0)$$ or
$$\sqrt{9+4a^2}+\sqrt{9+4b^2}\leq\sqrt{9+4(a+b)^2}+\sqrt{9}=3+\sqrt{13}.$$
Id est, it's enough to prove that
$$2(a^2+b^2)+3+\sqrt{13}\leq5+\sqrt{13}$$ or $$2(a^2+b^2)\leq2(a+b)^2,$$ which is obvious.
Done!
