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Using the Lagrange multipliers method, I have found that the maximum and minimum values of the function $$f(x,y)=xy$$ on the curve $$x^2-yx+y^2=1$$ are $$1 \ \text{at} \ (\pm1,\pm1) \ \ \ \text{and} \ \ -\frac{1}{3} \ \text{at} \ \big(\pm\frac{1}{\sqrt{3}},\mp\frac{1}{\sqrt{3}}\big) \ \ \text{respectively.}$$ Using this, prove that $$\Big|\frac{xy}{x^2-yx+y^2}\Big|\leq1 \ \ \ \forall(x,y)\neq0$$

I don't really know where to start. I thought of multiplying across as the inequality will be preserved, which yields $$|xy|\leq|x^2-yx+y^2|$$ then $$|xy|-|x^2-yx+y^2|\leq 0$$ At this point, I'm relying on my algebraic manipulation skills to yield something true. I'm not using the information above obtained by the Lagrange multipliers method.

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You can write $$-1 = \frac{-x^2+xy-y^2}{x^2-xy+y^2} \leq \frac{xy}{x^2-xy+y^2} = \frac{xy}{(x-y)^2+xy} \leq \frac{xy}{xy}=1.$$

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Equivalently, you prove: $(xy)^2 - (x^2-xy+y^2)^2 \le 0\iff (xy-x^2+xy-y^2)(xy+x^2-xy+y^2) \le 0\iff(2xy-x^2-y^2)(x^2+y^2) \le 0\iff -(x-y)^2(x^2+y^2) \le 0$ which is true.

We can show the max/min result without bothering Lagrange ! We have:

$(x-y)^2 \ge 0 \implies x^2-2xy+y^2 \ge 0 \implies x^2 -xy+y^2 \ge xy \implies 1 \ge xy\implies \text{max} = 1$

And observe $xy \ge -\dfrac{x^2-xy+y^2}{3}= -\dfrac{1}{3}\implies \text{min} = -\dfrac{1}{3}$ .

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  • $\begingroup$ This looks great. But does this utilise the results obtained by the Lagrange method? $\endgroup$ – user557493 Jun 15 '18 at 8:57
  • $\begingroup$ It is independent result and it need not use Lagrange Method. $\endgroup$ – DeepSea Jun 15 '18 at 8:59
  • $\begingroup$ Ideally I would wish to use the results of the Lagrange method, as this is specifically required in the question. But thank you for you perfectly valid result. Thumbs up $\endgroup$ – user557493 Jun 15 '18 at 9:01
  • $\begingroup$ DeepSea.Related: math.stackexchange.com/questions/2820423/… .Greetings $\endgroup$ – Peter Szilas Jun 15 '18 at 9:16
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Making a change of variables

$$ x = r\cos\theta\\ y = r\sin\theta $$

$$ x^2-x y+y^2= 1\rightarrow r^2(1-\sin\theta\cos\theta) = 1\rightarrow r^2 = \frac{1}{1-\sin\theta\cos\theta} $$

and then

$$ xy \equiv \frac{\sin\theta\cos\theta}{1+\sin\theta\cos\theta} = \frac{\sin(2\theta)}{2-\sin(2\theta)} $$

hence

$$ -\frac 13 \le x y \le 1 $$

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Correct me if wrong.

For $(x,y) \not = (0,0)$:

Denominator: $|x^2+y^2 -xy| \ge$

$ (x^2+y^2) - |xy|\ge$

$2|xy| -|xy| =|xy|.$

Used: $x^2+y^2 \ge 2|xy|$

Hence :

$|\dfrac{xy}{x^2+y^2 -xy}| \le 1.$

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Define: $g(x,y)=x^2+xy+y^2$ as you have already proven the extremizers for g(x,y)=1 we should try and turn the problem into one where g(x,y)=1. $$(1)\: f(ax,by)=abf(x,y)$$$$(2)\:g(ax,by)=a^2x^2+abxy+b^2y^2$$ Let $g(x,y)=c, c\in\Bbb R \setminus \{0\}$ $${\vert c\vert }f\left(\frac{x}{\sqrt {\vert{c}\vert}},\frac{y}{\sqrt{\vert c \vert }}\right)=f(x,y)$$$${\vert c\vert }g\left(\frac{x}{\sqrt{ \vert c \vert}},\frac{y}{\sqrt{ \vert c \vert}}\right)=g(x,y)$$ These formulas are found by subbing into c $(1)$ and $(2)$.$$\frac{f(x,y)}{g(x,y)}=\frac{f\left(\frac{x}{\sqrt {\vert{c}\vert}},\frac{y}{\sqrt{\vert c \vert }}\right)}{g\left(\frac{x}{\sqrt{ \vert c \vert}},\frac{y}{\sqrt{ \vert c \vert}}\right)}$$ by construction the denominator is 1 and the top is maximal at 1 and minimal at $\frac{-1}{3}$ hence the result follows $$\frac{|f(x,y)|}{|g(x,y)|}\leq1$$

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The lagrange function is $L(x,y,λ)=f(x,y)−λg(x,y)$ To take out the maximum of this function you (effectively) partially differentiated both R.H.S and L.H.S and found the the value of lambda for which L is maximum by equating these differentials to zero i.e $∂L=0=∂f−λ∂g$

. This gave you $L(+_-1,+_-1,1)$ Which would then (being global maximum by definition) be greater than equal to the function itself and hence the inequality will result.Hope this helps.

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