Need help on a proof from Spivak's Calculus

  • #1
CaptainAmerica17
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Homework Statement


I'm currently working through Spivak independently and have reached the problems at the end of ch. 1.
The problem is:
Prove that if [itex] 0 < a < b [/itex], then [tex] a < \sqrt{ab} < \frac{a+b}{2} < b [/tex]

Homework Equations



Spivak's properties P1 - P12

The Attempt at a Solution


I was thinking that if I do it by cases, I can first try to prove [itex] a < \sqrt{ab} [/itex] and then do [itex] \frac{a+b}{2} < b[/itex]. After that, I can use [itex] 0 < a < b [/itex] to bring it all together. To start with [itex] a < \sqrt{ab} [/itex], I have: If [itex] a > 0 [/itex] and [itex] b > 0 [/itex], then [itex] ab > 0 [/itex]. I'm not sure where to go from there or if my approach is even correct.
 

Answers and Replies

  • #2
fresh_42
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Try and solve it from behind: start with ##\sqrt{ab} \leq \dfrac{a+b}{2}## and square this equation, do some algebra, and finally write it down in reverse order.
 
  • #3
CaptainAmerica17
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Try and solve it from behind: start with ##\sqrt{ab} \leq \dfrac{a+b}{2}## and square this equation, do some algebra, and finally write it down in reverse order.
Thanks for the hint. It helped a bit.
 
  • #5
Ray Vickson
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Homework Statement


I'm currently working through Spivak independently and have reached the problems at the end of ch. 1.
The problem is:
Prove that if [itex] 0 < a < b [/itex], then [tex] a < \sqrt{ab} < \frac{a+b}{2} < b [/tex]

Homework Equations



Spivak's properties P1 - P12

The Attempt at a Solution


I was thinking that if I do it by cases, I can first try to prove [itex] a < \sqrt{ab} [/itex] and then do [itex] \frac{a+b}{2} < b[/itex]. After that, I can use [itex] 0 < a < b [/itex] to bring it all together. To start with [itex] a < \sqrt{ab} [/itex], I have: If [itex] a > 0 [/itex] and [itex] b > 0 [/itex], then [itex] ab > 0 [/itex]. I'm not sure where to go from there or if my approach is even correct.

(1) What do you get when you expand out ##(\sqrt{b}-\sqrt{a})^2\: ?##
(2) What can you do with the fact that ##0 < a < b \; \Rightarrow 0 < a^2 < ab \: ?##
 
  • #6
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Try and solve it from behind: start with ##\sqrt{ab} \leq \dfrac{a+b}{2}## and square this equation, do some algebra, and finally write it down in reverse order.
Did you mean "start with ##\sqrt{ab} \geq \dfrac{a+b}{2}## and square this inequality?
 
  • #7
mathwonk
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what happens if you multiply the inequality a < b by a? or if you let b = a+e where e>0? or almost anything. what did you try?
 
  • #8
fresh_42
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Did you mean "start with ##\sqrt{ab} \geq \dfrac{a+b}{2}## and square this inequality?
No. I meant to start with the claim, and transform it until a true statement is found. Afterwards, it has to be written in the correct order, i.e. reverse to what has been done. In this case, there are no losses between the steps due to the inequalities, and as ##0< a < b## the root - function in this case - also behaves nicely.

I haven't recommended to turn a proof upside down, I only said that starting at the end sometimes gives the right hints and shows the way to proceed.
 
  • #9
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No. I meant to start with the claim, and transform it until a true statement is found.
That's legitimate if all the steps are reversible, but squaring both sides is not one of them. However, since it is given that both a and b are positive, squaring both sides is reversible in this case.
 
  • #10
CaptainAmerica17
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No. I meant to start with the claim, and transform it until a true statement is found. Afterwards, it has to be written in the correct order, i.e. reverse to what has been done. In this case, there are no losses between the steps due to the inequalities, and as ##0< a < b## the root - function in this case - also behaves nicely.

I haven't recommended to turn a proof upside down, I only said that starting at the end sometimes gives the right hints and shows the way to proceed.
When I did that, I ended up with the statement [itex] 2 \leq a+b [/itex], which must be true given what's stated in the problem.
 
  • #11
fresh_42
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I think it is legitimate even if single steps aren't reversible. This - of course - won't result in a proof, as it is here the case, where we don't need an indirect procedure, but it might lead to insights where the critical points are. So to get ideas about how to prove something, it is often a good idea to play with the result. E.g. ##\varepsilon-## proofs can frequently be found this way, simply because we don't know how to set the variable at the existence quantifier at the start - it's filled in when it's obvious how, even if the formally correct version of the proof has it at the beginning.

However, I admit one has to know that such a procedure does not yield a proof - only hints. And if we're lucky, as in the example above, everything is reversible.
 
  • #12
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When I did that, I ended up with the statement [itex] 2 \leq a+b [/itex], which must be true given what's stated in the problem.
No, that's not necessarily true. For example, if a = 1/2 and b = 1, then the chain of inequalities given in post #1 is true.
 
  • #13
fresh_42
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When I did that, I ended up with the statement [itex] 2 \leq a+b [/itex], which must be true given what's stated in the problem.
I get ##4ab \leq (a+b)^2 \Longleftrightarrow 0 \leq (a-b)^2##. How did you arrive at ##2 \leq a+b\,?##
 
  • #14
CaptainAmerica17
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I think it is legitimate even if single steps aren't reversible. This - of course - won't result in a proof, as it is here the case, where we don't need an indirect procedure, but it might lead to insights where the critical points are. So to get ideas about how to prove something, it is often a good idea to play with the result. E.g. ##\varepsilon-## proofs can frequently be found this way, simply because we don't know how to set the variable at the existence quantifier at the start - it's filled in when it's obvious how, even if the formally correct version of the proof has it at the beginning.

However, I admit one has to know that such a procedure does not yield a proof - only hints. And if we're lucky, as in the example above, everything is reversible.

This is a skill I have been learning very slowly, but surely. It's been especially useful for some set theory proofs I've done recently.
 
  • #15
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I think it is legitimate even if single steps aren't reversible.
Maybe sometimes, but not in general.

For example, if we are given that a < b, by squaring both sides we might conclude that ##a^2 < b^2##. This conclusion would be false if a = -1 and b = 1/2, because we would be concluding that ##(-1)^2 = 1 < (\frac 1 2)^2 = \frac 1 4##.
 
  • #16
CaptainAmerica17
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I get ##4ab \leq (a+b)^2 \Longleftrightarrow 0 \leq (a-b)^2##. How did you arrive at ##2 \leq a+b\,?##
I started with [itex] \sqrt{ab} \leq \frac{a+b}{2} [/itex] and just went from there like you said in your original message.
 
  • #17
fresh_42
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Maybe sometimes, but not in general.

For example, if we are given that a < b, by squaring both sides we might conclude that ##a^2 < b^2##. This conclusion would be false if a = -1 and b = 1/2, because we would be concluding that ##(-1)^2 = 1 < (\frac 1 2)^2 = \frac 1 4##.
Yes. One has to be cautious. I remember I had a thesis on the desk with basically the following conclusion:
$$
a < b \Longrightarrow ax < bx
$$
Of course ##x## had been an entire expression. Too bad ##x## could have been negative!

And in the case above, I agree, the crucial part is to clearly state, why squaring does no harm to the equivalence. And in this sense, my hint had been far from being a proof. Just a hint how to get an idea.
 
  • #18
fresh_42
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I started with [itex] \sqrt{ab} \leq \frac{a+b}{2} [/itex] and just went from there like you said in your original message.
But taking the square yields ##ab \leq \dfrac{(a+b)^2}{4}##. But read my previous post, as to why this is an allowed operation in this case!
 
  • #19
CaptainAmerica17
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But taking the square yields ##ab \leq \dfrac{(a+b)^2}{4}##. But read my previous post, as to why this is an allowed operation in this case!

I think I took to many steps in the wrong direction. I just kept reducing. I see how you got what you did, I'm going to reassess.
 
  • #20
WWGD
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Just a comment that the lh side is the so-cold AGI or Arohmetric-Geometric inequality. The leftmost is the geometric mean and the center one is the Arithmetic mean. It holds true for n>1.
 
  • #21
WWGD
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Make that so-called and Arithmetic.
EarlyWinter getting to me.
 

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