# Inequality of Positive Real Numbers: Proving (1-1/x^2)(1-1/y^2)≥9 with x+y≤1

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• anemone
In summary, the purpose of proving (1-1/x^2)(1-1/y^2)≥9 with x+y≤1 is to show that the product of two positive real numbers, both less than or equal to 1, is always greater than or equal to 9. This inequality is significant because it demonstrates the relationship between two positive real numbers and their product, and it can be proven using basic algebraic principles and properties of inequalities. It can also be generalized to any two positive real numbers that satisfy the condition of x+y≤1. This inequality has various real-world applications in areas such as economics, physics, and engineering, and it can also be used in mathematical proofs and equations.
anemone
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Let ##x,\,y>0## and ##x+y \leq 1##.

Prove that ##(1-\frac{1}{x^2})(1-\frac{1}{y^2})\geq 9##.

DeBangis21, Lynch101, Crazy G and 3 others
Here's a partial proof without calculus:

Suppose ##x## has been chosen to be to be some positive value ##x_0##. From the condition

$$0 < y \leq 1 - x$$

the value of y is still indeterminate. However, since ##\frac{1}{y^2}## is always positive, the minimum of

$$(1 - \frac{1}{x_0^2})(1-\frac{1}{?^2})$$

must always occur when the maximum y "budget" is used. (The two minuses multiplied from the first and second terms will cancel in a double negative manner.)

Thus, we may reduce the possible answers to the possibilities for exactly the cases when

$$y = 1 - x$$

without any loss of solutions.

Because the expression is symmetric about ##x## and ##1-x##, a maximum or minimum must occur at the halfway mark ##x = \frac{1}{2} ##.

$$(1 - \frac{1}{\frac{1}{2}^2})(1-\frac{1}{\frac{1}{2}^2}) = 9$$

Numerical checking that setting x = 0.6 and x = 0.4 gives a higher value than 9 shows that this is a minimum point, therefore:

$$(1 - \frac{1}{x^2})(1-\frac{1}{y^2}) \geq 9$$

-------------------------------------------

There is still the possibility however that the curve has other minimums other than on its line of symmetry:

Can it be shown without calculus that this hypothetical case is impossible?

Last edited:
topsquark
James1238765 said:
Can it be shown without calculus that this hypothetical case is impossible?
After your first argument that ##x+ y =1##, try, without loss of generality, ##x = \frac 1 2 + z##, ##y = \frac 1 2 -z## and simplify the expression in ##z##.

topsquark
anemone said:
Let ##x,\,y>0## and ##x+y \leq 1##.

Prove that ##(1-\frac{1}{x^2})(1-\frac{1}{y^2})\geq 9##.
For convenience (so that each bracketed expression is positive - which simplifies the explanation slightly) rewrite the inequality as:$$f(x,y) = \left(\frac 1{x^2}-1\right)\left(\frac{1}{y^2}-1\right) \ge 9$$Factorise. Let ##A = (\frac 1x - 1),~B= (\frac 1x + 1), ~C = (\frac 1y -1), ~D = (\frac 1y +1)## then ##f(x,y) =ABCD##.
________________________________________

Stage 1. Show that ##f(x,y) \ge 9## for ##x,y>0## and ##x+y=1##

##y = 1-x##

##\begin {flalign*}
AC &= \left(\frac 1x - 1 \right) \left(\frac 1y -1 \right)\\
&= \left(\frac 1x - 1 \right)\left(\frac1{1-x }-1\right)\\
&= \left(\frac {1-x}x \right)\left(\frac {x}{1-x}\right)\\
&= 1
\end {flalign*}##

##\begin {flalign*}
BD &= \left(\frac 1x + 1 \right) \left(\frac1y +1 \right)\\
&=\frac 1{xy} + \frac 1x + \frac 1y + 1\\
&= \frac 1{xy} + \frac {x + y}{xy} + 1\\
&= \frac 1{xy} + \frac 1{xy} + 1\\
&= \frac 2{xy} + 1
\end {flalign*}##

##x+y= 1## so the arithmetic mean of ##x## and ##y## is ½. Using the AM-GM inequality:

##½ \ge \sqrt {xy}~## ⇒##xy \le ¼##

##BD \ge \frac 2¼+ 1## ⇒ ## BD \ge 9##.

##AC = 1## and ##BD \ge 9## ⇒ ##f(x,y) \ge 9##.
___________________________________________________________________________

Stage 2. Show ##f(x,y) \gt 9## for ##x,y>0## and ##x+y<1##

(A bit waffly.)

When ##x## increases, ##\frac 1{x^2}## decreases monotonically; so ##(\frac 1{x^2}-1)## also decreases monotonically.

Similarly, when ##y## increases, ##(\frac 1{y^2}-1)## decreases monotonically.

So an increase in ##x## or ##y## makes ##\left(\frac 1{x^2}-1\right)\left(\frac{1}{y^2}-1\right)## decrease monotonically.

When ##x+y<1##, we can increase ##x## or ##y## up to the point where ##x+y= 1##. The increase makes ##f(x,y)## decrease monotonically. But when ##x+y=1##, the smallest possible value of ##f(x,y)## is 9 (from Stage 1, above). So the value of ##f(x,y)## when ##x+y<1## must have been greater than 9.

chwala, Office_Shredder, dextercioby and 1 other person
Steve4Physics said:
For convenience (so that each bracketed expression is positive - which simplifies the explanation slightly) rewrite the inequality as:$$f(x,y) = \left(\frac 1{x^2}-1\right)\left(\frac{1}{y^2}-1\right) \ge 9$$Factorise. Let ##A = (\frac 1x - 1),~B= (\frac 1x + 1), ~C = (\frac 1y -1), ~D = (\frac 1y +1)## then ##f(x,y) =ABCD##.
________________________________________

Stage 1. Show that ##f(x,y) \ge 9## for ##x,y>0## and ##x+y=1##

##y = 1-x##

##\begin {flalign*}
AC &= \left(\frac 1x - 1 \right) \left(\frac 1y -1 \right)\\
&= \left(\frac 1x - 1 \right)\left(\frac1{1-x }-1\right)\\
&= \left(\frac {1-x}x \right)\left(\frac {x}{1-x}\right)\\
&= 1
\end {flalign*}##

##\begin {flalign*}
BD &= \left(\frac 1x + 1 \right) \left(\frac1y +1 \right)\\
&=\frac 1{xy} + \frac 1x + \frac 1y + 1\\
&= \frac 1{xy} + \frac {x + y}{xy} + 1\\
&= \frac 1{xy} + \frac 1{xy} + 1\\
&= \frac 2{xy} + 1
\end {flalign*}##

##x+y= 1## so the arithmetic mean of ##x## and ##y## is ½. Using the AM-GM inequality:

##½ \ge \sqrt {xy}~## ⇒##xy \le ¼##

##BD \ge \frac 2¼+ 1## ⇒ ## BD \ge 9##.

##AC = 1## and ##BD \ge 9## ⇒ ##f(x,y) \ge 9##.
___________________________________________________________________________

Stage 2. Show ##f(x,y) \gt 9## for ##x,y>0## and ##x+y<1##

(A bit waffly.)

When ##x## increases, ##\frac 1{x^2}## decreases monotonically; so ##(\frac 1{x^2}-1)## also decreases monotonically.

Similarly, when ##y## increases, ##(\frac 1{y^2}-1)## decreases monotonically.

So an increase in ##x## or ##y## makes ##\left(\frac 1{x^2}-1\right)\left(\frac{1}{y^2}-1\right)## decrease monotonically.

When ##x+y<1##, we can increase ##x## or ##y## up to the point where ##x+y= 1##. The increase makes ##f(x,y)## decrease monotonically. But when ##x+y=1##, the smallest possible value of ##f(x,y)## is 9 (from Stage 1, above). So the value of ##f(x,y)## when ##x+y<1## must have been greater than 9.
That's too complicated!

If ##x + y = 1##, then:
$$f(x, y) = \bigg (\frac 1 {x^2} -1 \bigg )\bigg (\frac 1 {y^2} -1 \bigg ) = \frac{(1 + x)(1 +y)}{xy}$$Then, with ##x = \frac 1 2 + z, \ y = \frac 1 2 -z, \ 0 \le z < \frac 1 2##$$f(x, y) = 1 + \frac{8}{1-4z^2}$$

[\SPOILER]

chwala, Office_Shredder, dextercioby and 3 others
PeroK said:
That's too complicated!
Alas, I couldn't come up with anything simpler! (Also, my answer is quite long because, for clarity, I gave all the low-level steps.)

chwala, Theia, PeroK and 1 other person
PeroK said:
If ##x + y = 1##, then:
$$f(x, y) = \bigg (\frac 1 {x^2} -1 \bigg )\bigg (\frac 1 {y^2} -1 \bigg ) = \frac{(1 + x)(1 +y)}{xy}$$Then, with ##x = \frac 1 2 + z, \ y = \frac 1 2 -z, \ 0 \le z < \frac 1 2##$$f(x, y) = 1 + \frac{8}{1-4z^2}$$

[\SPOILER]
That's a very nice way to deal with the case where ##x+y=1##. But note that the requirement is for ##x+y \le 1##. For example ##x=0.8, y=0.1##. In that part of the domain ##x = \frac 12 + z, \ y = \frac 12 -z## can’t be used.

Steve4Physics said:
That's a very nice way to deal with the case where ##x+y=1##.
That was already established by @James1238765. If we fix ##x##, then the minimum occurs with the largest ##y##.

My solution picks up from that preliminary observation.

See post #3.

Last edited:
Steve4Physics
James1238765 said:
There is still the possibility however that the curve has other minimums other than on its line of symmetry:

View attachment 320169

Can it be shown without calculus that this hypothetical case is impossible?
The case shown in this drawing (symmetric minimums around the main minimum) must be impossible because there simply aren't enough powers to make all those inflection points.

@bob012345 i was thinking along those lines, but could not recall the theorem that links the number of inflection points to the degree of the powers of ##\frac{1}{x^n}##. Could anyone state that particular theorem, please?

bob012345
James1238765 said:
@bob012345 i was thinking along those lines, but could not recall the theorem that links the number of inflection points to the degree of the powers of ##\frac{1}{x^n}##. Could anyone state that particular theorem, please?
I should have said maxima and minima which is what I meant, not inflection points which are different. Anyway, it's going to be one less than the degree of the polynomial since one takes the derivative and sets it equal to zero to find the extrema. For example you would need a sixth degree polynomial to get the five extrema in your drawing but as I understand it not all sixth degree polynomials will have that many extrema.

https://www.mathsisfun.com/calculus/inflection-points.html

This isn't a proof but if we let the function be greater than some unknown ##n##
$$f(x,y)=\left(1-\ \frac{1}{x^{2}}\right)\left(1-\frac{1}{y^{2}}\right)>=n$$ we can also write it in this form
$$x^2+y^2 <=1-(n-1)x^2y^2$$ when ##n=1## it is a unit circle but we can also see that this curve will intersect a circle ##x^2+y^2=\frac{1}{2}## on the edge of our domain and the line ##y=x## at point {##\frac{1}{2},\frac{1}{2}##}.
$$x^2+y^2 =\frac{1}{2}=1-(n-1)\frac{1}{16}$$ giving ##n=9##

Graphically, letting it plot in all quadrants because it looks nice and plotting it over our domain of interest it shows that the curve doesn't contain all the points for ##n>9##

for n=1;

for n=9;

for n=18;

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