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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.

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closed as off-topic by Saad, David G. Stork, TheSimpliFire, Cesareo, Eevee Trainer Apr 19 at 0:27

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    $\begingroup$ What does the first equation have to do with the second??? $\endgroup$ – David G. Stork Apr 14 at 1:29
  • $\begingroup$ Obvious that $(x,y)$ in the inequality must satisfy the equation. $\endgroup$ – Piquito Apr 14 at 2:14
  • $\begingroup$ Thanks, David G. Stork, for the edition (my English is deficient). I will delete this problem but if you want you can see the link given to Will Jagy below. $\endgroup$ – Piquito Apr 14 at 2:44
  • $\begingroup$ Yes, $\max\limits_{\sqrt{x^2-3x}+\sqrt{y^2-3y}=1}(x^2+y^2-2(x+y))=\frac{11+\sqrt{13}}{2}.$ $\endgroup$ – Michael Rozenberg Apr 14 at 5:25
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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!

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    $\begingroup$ Your classic "Done!". You are strong, Michael. $\endgroup$ – Piquito Apr 15 at 13:01

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