Inequality: Prove for every two positive integers a and b that (a+b)(1a+1b)≥4

Roger.Robert

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Prove for every two positive integers a and b that \(\displaystyle (a+b)(\frac{1}{a}+\frac{1}{b}) \geq 4\)
By multiplying I obtained : \(\displaystyle \frac{a^{2}+b^{2}}{ab} \geq 2\)
And this is where I got stuck.
 
Of course, you can't, strictly speaking, prove a statement by assuming it is true!

What you can do is start from the \(\displaystyle (a- b)^2= a^2- 2ab+ b^2\ge 0\) and work back to the statement you are trying to prove.

(Yes, there is a method, sometimes called "synthetic proof", in which one starts from the statement one is trying to prove, here \(\displaystyle (a- b)(\frac{1}{a}+ \frac{1}{b}\ge 4\), and deriving an "obviously true" statement, "\(\displaystyle (a- b)^2\ge 0\)". That's valid as long as every step is reversible. The "real proof" would start at the statement \(\displaystyle (a- b)^2\ge 0\) and show \(\displaystyle (a- b)(\frac{1}{a}+ \frac{1}{b}\ge 4\).)

(Do you see why it is important that a and b be positive?)
 
Heh. Didn't realize my proof was synthetic.

Continuing from where Roger.Robert left off, I set the inequality to zero and factored the left-hand side.

There's only two cases to consider: a=b or a≠b :cool:
 
Isn't this a trivial proof as \(\displaystyle (a-b)^{2} \geq 0\) is true for any value of a or b?
 
Actually, a and b are not just "any value". The given inequality is not true, if either a or b is zero or negative.

The point is not to prove (a-b)^2 ≥ 0 (although you need to).

We're using the fact that we can show it's true, in proving the given inequality.

It's straightforward to prove (a-b)^2 ≥ 0 for two positive numbers a and b (take cases).

As Halls mentioned, you can reverse the steps.

That is, once you've shown that (a-b)^2 ≥ 0 is true, show that it's equivalent to the given inequality.
 
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