- #1
anemone
Gold Member
MHB
POTW Director
- 3,883
- 115
Let ##x,\,y>0## and ##x+y \leq 1##.
Prove that ##(1-\frac{1}{x^2})(1-\frac{1}{y^2})\geq 9##.
Prove that ##(1-\frac{1}{x^2})(1-\frac{1}{y^2})\geq 9##.
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##.James1238765 said:Can it be shown without calculus that this hypothetical case is impossible?
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##.
That's too complicated!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.
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.)PeroK said:That's too complicated!
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.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 was already established by @James1238765. If we fix ##x##, then the minimum occurs with the largest ##y##.Steve4Physics said:That's a very nice way to deal with the case where ##x+y=1##.
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.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?
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.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?