# 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: Note by Oussama Jaber
6 years, 11 months ago

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$2$ 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 $x(a-b)>0$]

- 6 years, 11 months ago

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?

- 6 years, 11 months ago

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

- 6 years, 11 months ago