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

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The discussion revolves around proving the inequality (1/a) + (1/b) ≥ (2/(a+b)) for positive values of a and b. The initial attempt leads to the expression (a+b)(a-b), which is questioned for its validity since it can yield negative results when a and b are not both positive. Participants highlight that the steps taken in the proof involve assumptions that could invalidate the inequality if a or b were negative. The need to maintain the direction of inequalities during multiplication is emphasized to ensure the proof's correctness. Ultimately, the conclusion suggests that a more careful approach is necessary to align with the textbook's requirements for inequalities.
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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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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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