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
4 years, 5 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 $$x(a-b)>0$$]

- 4 years, 5 months ago

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

- 4 years, 5 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?

- 4 years, 5 months ago