Difference between revisions of "1969 Canadian MO Problems/Problem 2"
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<math>\sqrt{c+1}-\sqrt c=\frac{(\sqrt{c+1})^2-(\sqrt c)^2}{\sqrt{c+1}+\sqrt{c}}=\frac1{\sqrt{c+1}+\sqrt{c}}.</math> | <math>\sqrt{c+1}-\sqrt c=\frac{(\sqrt{c+1})^2-(\sqrt c)^2}{\sqrt{c+1}+\sqrt{c}}=\frac1{\sqrt{c+1}+\sqrt{c}}.</math> | ||
| − | Similarly, <math>\sqrt c-\sqrt{c-1}=\frac{1}{\sqrt c | + | 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>. |
{{Old CanadaMO box|num-b=1|num-a=3|year=1969}} | {{Old CanadaMO box|num-b=1|num-a=3|year=1969}} | ||
, 
is greater for any 
.
by its conjugate,
. We know that 
for all positive 
, so 
.
