Difference between revisions of "1969 Canadian MO Problems/Problem 2"
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Determine which of the two numbers <math>\sqrt{c+1}-\sqrt{c}</math>, <math>\sqrt{c}-\sqrt{c-1}</math> is greater for any <math>c\ge 1</math>. | Determine which of the two numbers <math>\sqrt{c+1}-\sqrt{c}</math>, <math>\sqrt{c}-\sqrt{c-1}</math> is greater for any <math>c\ge 1</math>. | ||
| − | == Solution == | + | == Solution 1 == |
Multiplying and dividing <math>\sqrt{c+1}-\sqrt c</math> by its conjugate, | Multiplying and dividing <math>\sqrt{c+1}-\sqrt c</math> by its conjugate, | ||
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Similarly, <math>\sqrt c-\sqrt{c-1}=\frac{1}{\sqrt c+\sqrt{c-1}}</math>. We know that <math>\frac1{\sqrt{c+1}+\sqrt{c}}<\frac{1}{\sqrt c+\sqrt{c-1}}</math> for all positive <math>c</math>, so <math>\sqrt{c+1}-\sqrt c <\sqrt c-\sqrt{c-1}</math>. | Similarly, <math>\sqrt c-\sqrt{c-1}=\frac{1}{\sqrt c+\sqrt{c-1}}</math>. We know that <math>\frac1{\sqrt{c+1}+\sqrt{c}}<\frac{1}{\sqrt c+\sqrt{c-1}}</math> for all positive <math>c</math>, so <math>\sqrt{c+1}-\sqrt c <\sqrt c-\sqrt{c-1}</math>. | ||
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| + | == Solution 2 == | ||
| + | Considering the derivative of <math>f(x)=\sqrt{x+1}-\sqrt{x}</math>. | ||
| + | |||
| + | We have <math>f'(x)=\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{x}}</math>. Putting under a common denominator, we can see that the top will be negative. | ||
| + | |||
| + | Thus <math>\boxed{\sqrt{c}-\sqrt{c-1}}</math> is greater. | ||
| + | |||
| + | ~hastapasta | ||
{{Old CanadaMO box|num-b=1|num-a=3|year=1969}} | {{Old CanadaMO box|num-b=1|num-a=3|year=1969}} | ||
Revision as of 20:45, 19 December 2022
Problem
Determine which of the two numbers 
, 
is greater for any 
.
Solution 1
Multiplying and dividing 
by its conjugate,
Similarly, 
. We know that 
for all positive 
, so 
.
Solution 2
Considering the derivative of 
.
We have 
. Putting under a common denominator, we can see that the top will be negative.
Thus 
is greater.
~hastapasta
| 1969 Canadian MO (Problems) | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 3 |
