Difference between revisions of "1971 Canadian MO Problems/Problem 2"

(Created page with "== Problem == Let <math>x</math> and <math>y</math> be positive real numbers such that <math>x+y=1</math>. Show that <math>\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\ge...")
 
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which is true since by AM-GM, we get: <cmath>x^2 + y^2 \ge 2xy</cmath> <cmath>x^2 + 2xy + y^2 \ge 4xy</cmath> <cmath> (x+y)^2 \ge 4xy</cmath> and we are given that <math> x + y = 1</math>, so <cmath> (x+y)^2 \ge 4xy \Rightarrow 1 \ge 4xy </cmath>
 
which is true since by AM-GM, we get: <cmath>x^2 + y^2 \ge 2xy</cmath> <cmath>x^2 + 2xy + y^2 \ge 4xy</cmath> <cmath> (x+y)^2 \ge 4xy</cmath> and we are given that <math> x + y = 1</math>, so <cmath> (x+y)^2 \ge 4xy \Rightarrow 1 \ge 4xy </cmath>
  
{{Old CanadaMO box|num-b=3|num-a=5|year=1971}}
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{{Old CanadaMO box|num-b=1|num-a=3|year=1971}}
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 22:24, 13 December 2011

Problem

Let $x$ and $y$ be positive real numbers such that $x+y=1$. Show that $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\ge 9$.

Solution

\[\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \ge 9\] \[1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \ge 9\] \[\frac{xy+x+y+1}{xy}\ge 9\] \[\frac{xy+2}{xy}\ge 9\] \[1 + \frac{2}{xy}\ge 9\] \[\frac{2}{xy} \ge 8\] \[1\ge 4xy\] which is true since by AM-GM, we get: \[x^2 + y^2 \ge 2xy\] \[x^2 + 2xy + y^2 \ge 4xy\] \[(x+y)^2 \ge 4xy\] and we are given that $x + y = 1$, so \[(x+y)^2 \ge 4xy \Rightarrow 1 \ge 4xy\]

1971 Canadian MO (Problems)
Preceded by
Problem 1
1 2 3 4 5 6 7 8 Followed by
Problem 3