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

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.

@@ Line 19: / Line 19: @@
 <cmath>3\cdot 3</cmath>
 <cmath>9</cmath>
-Thus the minimum is <math>\boxed{9}</math>
+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]]

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

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