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Let $x,y \in \mathbb R$ and $$(1+x+x^2)(1+y+y^2)=2x^2y^2-1$$ and assume that $2<x<y$. I want to algebraically show that $xy<16$ but I cannot get anywhere. I have tried substitution and nothing seems to help. Graphically I know that it is try because I can simply graph and check visually but I want an actual proof if any one can offer any hints or proof thank you.

Jean Marie
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thestar
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    Hmm. I get $xy>7$. – amsmath Mar 07 '21 at 19:58
  • That is related so how to show then that $7<xy<16$ ? – thestar Mar 07 '21 at 19:59
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    I'm not sure if this will be useful, but you can say that $xy = \left(\frac{1}{2}((1+x+x^2)(1+y+y^2)+1)\right)^{1/2}$. – DMcMor Mar 07 '21 at 20:00
  • We can say that yes @DMcMor – thestar Mar 07 '21 at 20:01
  • @MartinR If $x\to +\infty,$ then $y$ becomes smaller and smaller, and vice versa. This does not contradict anything in OP. – Allawonder Mar 07 '21 at 20:46
  • That's what it looks like: https://www.wolframalpha.com/input/?i=Plot%5B%7B-%28%281+%2B+x+%2B+x%5E2%29%2F%282+%281+%2B+x+-+x%5E2%29%29%29+%2B+++++Sqrt%5B%282+%2B+x+%2B+x%5E2%29%2F%28+++++1+%2B+x+-+x%5E2%29+%2B+%28%281+%2B+x+%2B+x%5E2%29%2F%282+%281+%2B+x+-+x%5E2%29%29%29%5E2%5D%2C+x%2C+16%2Fx%7D%2C+%7Bx%2C++++2%2C+2.3%7D%5D – amsmath Mar 07 '21 at 20:46
  • @Allawonder: Ignore that, it was a typo when entering the formula. – Martin R Mar 07 '21 at 20:51

2 Answers2

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$$(1+x+x^2)(1+y+y^2)=2x^2y^2-1\tag{1}$$ $2<x<y$.

Plug $x=2$ in $(1)$ $$7 \left(y^2+y+1\right)=8 y^2-1\to y=-1;\;y=8$$ $(2,8)$ is a point where the curve has its maximum because in that point the curve $(1)$ is decreasing $$y'=\frac{-2 x y^2+2 x y+2 x+y^2+y+1}{2 x^2 y-x^2-2 x y-x-2 y-1}$$ as $y'(2,8)=-\frac{49}{3}<0$. It can be proved that the curve is decreasing in the whole region $2<x<y$.

Thus the maximum value of $xy$ is $16$.

Hope this helps

$$...$$ enter image description here

Raffaele
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Here is a proof which, I am conscious of that, is graphical but has the merit to transform the issue into more natural variables for the problem at hand, i.e., by taking:

$$\begin{cases}S=x+y\\ P=xy\end{cases} \ \ \iff \ \ \begin{cases}x&=&\frac12(S-\sqrt{S^2-4P})\\y&=&\frac12(S+\sqrt{S^2-4P})\end{cases}\tag{1}$$

(think to relationship $t^2-St+P=0$).

Indeed, with this change of variables, the problem is easily shown to be equivalent (due to symmetry between variables $x$ and $y$) to:

$$\text{knowing that} \ \ -P^2+PS+S^2-P+S+2=0 \tag{2}$$

$$\text{show that} \ \ P<16$$

taking into account constraint $x>2$ (see (1)) and the implied constraint $S^2-4P \ge 0$.

Now, have a look at fig. 1 below with coordinates axes $(S,P)$:

enter image description here

Fig. 1: Only the region defined by $(S>4,P>4)$ makes sense. We have extended the "scene" in order to better understand the hyperbolic curve given by equ. (2).

Point $(S,P)$ must

  • belong to hyperbola (in blue) accounting for constraint (2).

  • be below parabola (P) (in black) with equation $4P=S^2$ and

  • be above the red line accounting for condition $x>2.$

The limit point is at the intersection of the blue and red curves (small square) with ordinate $S=16$ as desired.

I am convinced that one can deduce from this graphical representation a fully analytic proof.

Jean Marie
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