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)
Easy Math Editor
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2 is simple. For positive numbers a and b, 2a+b≥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≥0
⇒(a+b)2−4ab≥0
And after a little bit of work, 2a+b≥ab.
For 1, each one of the following can be true:
ba=b+xa+x [happens when x=0 or a=b]
ba<b+xa+x [happens when x(a−b)<0]
ba>b+xa+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
2a+b−ab=2a+b−2ab=2(a−b)2≥0
therefore, 2a+b≥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?
sorry I missed one condition from problem 1.... x>0