Question

Prove that for every two positive integers and and b that (a+b)(1/a + 1/b) is greater than or equal to 4

Answer 1

I've tried manipulating the equation around a bit, but nothing I work out seems intuitively obvious. Is there an easy way to prove this I'm just not seeing?

Answer 2

Yeah both assumptions are wrong, since we could have a rational number less than 1 depending on how you pick a and b... But I wouldn't throw away all your work because it's wrong otherwise it doesn't help us get to any real understanding.

Answer 3

\(\large \begin{align} \color{black}{(a+b)\geq 2--------\color{red}{(1)}\hspace{.33em}\\~\\
\normalsize \text{also }\hspace{.33em}\\~\\
\dfrac{1}{a}+\dfrac{1}{b}\leq 2--------\color{red}{(2)}\hspace{.33em}\\~\\
}\end{align}\)
are these right

Answer 4

Yeah those are definitely true statements =D

Answer 5

haha

Answer 6

\[(a-b)^2 = (a+b)^2 - 4ab \ge 0 \implies (a+b)(a+b) \ge 4ab\]

divide \(ab\) both sides

Answer 7

Notice \(ab\gt 0\) is required for dividing \(ab\) both sides w/o bothering about flipping the inequality

Answer 8

Just a comment, but I think 'rabbit out of the hat' proofs should be avoided, if at all possible. (Sometimes, this is sadly impossible). By which I mean, I think it would probably be more instructive if you'd started at (a+b)(1/a + 1/b) >= 4, then arrived at (a+b)^2 >= 4ab iff. (a+b)^2 - ab >=0 ... etc, rather than start with (a-b)^2 + 4ab out of seemingly nowhere, and getting back to the original inequality.

Might just be a personal preference.

Answer 9

@INewton I don't like proofs that start with the given statement because it is easy to make errors. Suppose you want to prove \(a \ge b\), below is a very bad proof strategy :

\[a =b \implies x = y \cdots \implies 1 =1 \]

as the implications need not be true in the other direction. It is always good to start with an identity like \(1=1\) and arrive at the statement you're trying to prove.

Answer 10

But we're not writing proofs, we're trying to explain how to solve problems. How did you solve the problem? Working from what was given, or realising it related to (a-b)^2 and working backwards? Granted, sometimes the latter may be the obvious route, but usually the former seems far more natural to me.

As for writing the actual proof, the implications are almost invariably if and only if statements. Even if they were not, it's perfectly valid to write (say):

\[(a+b) \frac{a+b}{ab} = \frac{(a+b)^2}{ab} \geq 4 \underbrace{\impliedby}_{ab > 0} (a+b)^2 \geq 4ab \impliedby (a-b)^2 \geq 0 \]
(Of course, you'd normally write iff. in each case here, but it's not unimaginable that sometimes iff would not hold)