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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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== Solution ==
 
== Solution ==
 +
===Solution 1===
 
<cmath>\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \ge 9</cmath>
 
<cmath>\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \ge 9</cmath>
 
<cmath>1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \ge 9 </cmath>
 
<cmath>1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \ge 9 </cmath>
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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}}
+
===Solution 2===
 +
Let <math>f(x)=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{1-x}\right)</math>. Since we want to find the minimum of the function over the [[interval]] <math>(0,1)</math>, we can take the [[derivative]]. Using the [[product rule]], we get <math>\frac{2(2x-1)}{x^2(1-x)^2}</math>. Since we want this value to be zero, the numerator must be zero. But the only root of the equation <math>2(2x-1)</math> is <math>1/2</math>, and so plugging the answer back in we have
 +
<cmath>(1+\frac{1}{\frac{1}{2}})(1+\frac{1}{\frac{1}{2}})</cmath>
 +
<cmath>(1+2)(1+2)</cmath>
 +
<cmath>3\cdot 3</cmath>
 +
<cmath>9</cmath>
 +
Thus the minimum is <math>\boxed{9}</math> and we are done.
 +
 
 +
== See Also ==
 +
{{Old CanadaMO box|num-b=1|num-a=3|year=1971}}
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Latest revision as of 16:31, 16 January 2023

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

Solution 1

\[\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \ge 9\] \[1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \ge 9\] \[\frac{xy+x+y+1}{xy}\ge 9\] \[\frac{xy+2}{xy}\ge 9\] \[1 + \frac{2}{xy}\ge 9\] \[\frac{2}{xy} \ge 8\] \[1\ge 4xy\] 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 \[(x+y)^2 \ge 4xy \Rightarrow 1 \ge 4xy\]

Solution 2

Let $f(x)=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{1-x}\right)$. Since we want to find the minimum of the function over the interval $(0,1)$, we can take the derivative. Using the product rule, we get $\frac{2(2x-1)}{x^2(1-x)^2}$. Since we want this value to be zero, the numerator must be zero. But the only root of the equation $2(2x-1)$ is $1/2$, and so plugging the answer back in we have \[(1+\frac{1}{\frac{1}{2}})(1+\frac{1}{\frac{1}{2}})\] \[(1+2)(1+2)\] \[3\cdot 3\] \[9\] Thus the minimum is $\boxed{9}$ and we are done.

See Also

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