Solving Abstract Inequalities: Proving (a+b)(a-b) = 0 for Positive a and b

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Homework Help Overview

The discussion revolves around proving an inequality involving positive variables a and b, specifically showing that \(\frac{1}{a}+\frac{1}{b}\geq\frac{2}{a+b}\). Participants are examining the implications of manipulating the expression \((a+b)(a-b)\) and its relationship to the conditions set by a textbook.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the validity of steps taken in the manipulation of the inequality, questioning the assumptions made when multiplying by expressions that could be negative. There is also a focus on whether the derived expression \((a+b)(a-b)\) holds true under the conditions specified.

Discussion Status

Some participants have raised concerns about the validity of the steps taken in the original poster's attempt, particularly regarding the implications of multiplying by potentially negative quantities. There is ongoing exploration of the conditions under which the derived inequalities hold.

Contextual Notes

Participants note that the original inequality must hold for all real numbers, which adds complexity to the discussion. The original poster expresses uncertainty about their solution's alignment with the textbook's requirements.

odolwa99
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Homework Statement



The final answer I have of (a+b)(a-b) does not appear to fit the textbook's required "results of inequalities which hold true for all real no.s", i.e. either: 1. (a)^2 or (a-b)^2 or 2. -(a+b)^2. Can anyone confirm if I have solved this correctly, in line with the conditions the book has described above?

Many thanks.

Q. If a > 0 & b > 0, show that: \frac{1}{a}+\frac{1}{b}\geq\frac{2}{a+b}

Homework Equations

The Attempt at a Solution



Attempt: if \frac{a+b}{ab}\geq\frac{2}{a+b}
if a+b\geq\frac{2ab}{a+b}
if (a+b)(a+b)\geq2ab
if a^2+2ab+b^2-2ab\geq0
if (a+b)(a-b)\geq0
 
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It should be obvious that (a+ b)(a- b) is NOT always positive: take a= 1, b= 2 so that (a+ b)(a- b)= (3)(-1)= -3.
 
In which case I should conclude with:
if a^2+b^2\geq0?
 
odolwa99 said:

Homework Statement



The final answer I have of (a+b)(a-b) does not appear to fit the textbook's required "results of inequalities which hold true for all real no.s", i.e. either: 1. (a)^2 or (a-b)^2 or 2. -(a+b)^2. Can anyone confirm if I have solved this correctly, in line with the conditions the book has described above?

Many thanks.

Q. If a > 0 & b > 0, show that: \frac{1}{a}+\frac{1}{b}\geq\frac{2}{a+b}


Homework Equations




The Attempt at a Solution



Attempt: if \frac{a+b}{ab}\geq\frac{2}{a+b}
if a+b\geq\frac{2ab}{a+b}
if (a+b)(a+b)\geq2ab
if a^2+2ab+b^2-2ab\geq0
if (a+b)(a-b)\geq0

Why do you have "if" starting each line?

You have done several things that aren't always valid.
1. You got to the 2nd step by multiplying both sides of the original inequality by ab. If a and b have opposite signs, the inequality direction in step 2 needs to switch.
2. To get to step 3, you multiplied by a + b. If a + b < 0, the inequality direction needs to switch.

You haven't taken this into consideration, so what you ended with doesn't necessarily follow what you started with.
 
To account for the 'if's', I'll post an image of an example from the book where I got this question from (apologies for the slightly fuzzy image).
As for points 1. & 2., I guess I will need to keep the clutter away from the RHS of the equation, so the inequality remains true, and I avoid having to change signs.
 

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