Comparing two real nubmers

I have been struggling for 2 days now trying to solve these Math problems. I hope someone could solve these in the next few minutes or so.
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)

Note by Oussama Jaber
6 years, 11 months ago

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22 is simple. For positive numbers aa and bb, a+b2ab\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 (ab)20(a-b)^2\geq 0

(a+b)24ab0\Rightarrow (a+b)^2 -4ab\geq 0

And after a little bit of work, a+b2ab\frac{a+b}{2}\geq \sqrt{ab}.

For 11, each one of the following can be true:

ab=a+xb+x\frac{a}{b}=\frac{a+x}{b+x} [happens when x=0x=0 or a=ba=b]

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

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

Mursalin Habib - 6 years, 11 months ago

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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

therefore, a+b2ab\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?

Oussama Jaber - 6 years, 11 months ago

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

Oussama Jaber - 6 years, 11 months ago

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