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

[odolwa99]

Homework Statement

The final answer I have of [itex](a+b)(a-b)[/itex] does not appear to fit the textbook's required "results of inequalities which hold true for all real no.s", i.e. either: 1. [itex](a)^2[/itex] or [itex](a-b)^2[/itex] or 2. [itex]-(a+b)^2[/itex]. 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: [itex]\frac{1}{a}+\frac{1}{b}\geq\frac{2}{a+b}[/itex]

Homework Equations

The Attempt at a Solution

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

 

[HallsofIvy]

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.

 

[odolwa99]

In which case I should conclude with:
if [itex]a^2+b^2\geq0[/itex]?

 

[Mark44]

Homework Statement

The final answer I have of [itex](a+b)(a-b)[/itex] does not appear to fit the textbook's required "results of inequalities which hold true for all real no.s", i.e. either: 1. [itex](a)^2[/itex] or [itex](a-b)^2[/itex] or 2. [itex]-(a+b)^2[/itex]. 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: [itex]\frac{1}{a}+\frac{1}{b}\geq\frac{2}{a+b}[/itex]

Homework Equations

The Attempt at a Solution

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

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.

 

[odolwa99]

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.