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
5 years, 8 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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 - 5 years, 8 months ago

Log in to reply

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

a+b2ab=a+b2ab2=(ab)220\frac{a+b}{2}-\sqrt{ab}=\frac{a+b-2\sqrt{ab}}{2}=\frac{(\sqrt{a}-\sqrt{b})^2}{2}\geq0
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 - 5 years, 8 months ago

Log in to reply

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

Oussama Jaber - 5 years, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...