# Finding a better bound in an inequality [closed]

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.

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

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• What does the first equation have to do with the second??? – David G. Stork Apr 14 at 1:29
• Obvious that $(x,y)$ in the inequality must satisfy the equation. – Piquito Apr 14 at 2:14
• 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. – Piquito Apr 14 at 2:44
• Yes, $\max\limits_{\sqrt{x^2-3x}+\sqrt{y^2-3y}=1}(x^2+y^2-2(x+y))=\frac{11+\sqrt{13}}{2}.$ – Michael Rozenberg Apr 14 at 5:25

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!

• Your classic "Done!". You are strong, Michael. – Piquito Apr 15 at 13:01