Subject: Order on R

1) a and b being two non zero numbers and x>0, compare:

a/b and (a+x)/(b+x)

2) a and b being two strictly positive numbers, compare:

(a+b)/ 2 and radical(ab)

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## Comments

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TopNewest\(2\) is simple. For positive numbers \(a\) and \(b\), \(\frac{a+b}{2}\geq \sqrt{ab}\). This is the two variable case of the Arithmetic Mean-Geometric Mean inequality. Here's a simple proof of the two variable case:

We know that \((a-b)^2\geq 0\)

\(\Rightarrow (a+b)^2 -4ab\geq 0\)

And after a little bit of work, \(\frac{a+b}{2}\geq \sqrt{ab}\).

For \(1\), each one of the following can be true:

\(\frac{a}{b}=\frac{a+x}{b+x}\) [happens when \(x=0\) or \(a=b\)]

\(\frac{a}{b}<\frac{a+x}{b+x}\) [happens when \(x(a-b)<0\)]

\(\frac{a}{b}>\frac{a+x}{b+x}\) [happens when \(x(a-b)>0\)]

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Thank you for your reply, We did not take about the Arithmetic Mean- Geometric Mean inequality. The only method we took when comparing two numbers are:

Compare their difference, compare their squares, compare their radicals ...ect

I managed to to solve 2 by comparing their difference. Here is how I solved it.................................................... a and b are strictly positive, therefore a >0 and b>0

\(\frac{a+b}{2}-\sqrt{ab}=\frac{a+b-2\sqrt{ab}}{2}=\frac{(\sqrt{a}-\sqrt{b})^2}{2}\geq0\)

therefore, \(\frac{a+b}{2}\geq\sqrt{ab}\)

But for the problem 1, I knew that there is more than one answer because I plugged in numbers to check for comparison and get different comparison each time I plug in a different number.

But how can use the methods we learned to prove these different comparisons?

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sorry I missed one condition from problem 1.... x>0

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