AM-GM Inequality!

Algebra Level 3

True or False

Positive real numbers $x$ and $y$ are such that $x+y < 2$ . By AM-GM inequality , we have:

$\large xy+\frac{1}{xy}\ge2\sqrt{xy\left(\frac{1}{xy}\right)}=2$

Then is it true that $xy+\dfrac{1}{xy}$ has a minimum value of 2?

True False

2 solutions

Kushal Bose
May 30, 2017

if the minimum value is True then equality holds.The condition of holding equality in AM-GM is terms will be equal

Then $xy=1/xy => (xy)^2=1 =>xy=1$ as $x.y >0$

So, $y=1/x$ then $x+1/x <2$ It contradicts AM-GM bcoz $x+1/x \geq 2 \sqrt{x.1/x}=2$

So, what would the minimum value be then?

Calvin Lin Staff - 4 years ago

Never think about this

let me try

Kushal Bose - 4 years ago

I suppose the minimum value would be the least real number greater than $2$ , but since there is no such number we would have to conclude that there is no distinct minimum value.

Brian Charlesworth - 4 years ago

@Brian Charlesworth – Indeed! There's an infinium, but no supremum.

This is essentially demonstrated by the above argument, but might need 1-2 lines to fill in the details.

Calvin Lin Staff - 4 years ago

@Calvin Lin – If I want to demonstrate by geometrically then we need to find points under the straight line x+y=2 ,x>0,y>0 .xy means the area of rectangle formed by length and breadth x and y respectively.Now consider a rectangular hyperbola xy=1 which will touch at single point of the line x+y=1.We have selected x and now 1/x is the point on the hyperbola and similarly for 1/y.Then we get two separate rectangle.The minimum of the area will be when this two will merge and this will be possible when we select the point on the hyperbola.Then x and y will be (1,1).But we need to take x+y<2 so total area will be $lim x \to 2^+$ .

The minimum will be a limiting value of greater than 2.

Kushal Bose - 4 years ago

Sir can we use linear programming method in this ??

Ayush Sharma - 4 years ago

Chew-Seong Cheong
May 31, 2017

Equality of the inequality $xy+\dfrac{1}{xy}\ge2\sqrt{xy\left(\dfrac{1}{xy}\right)}=2$ occurs when $xy = \dfrac 1{xy}$ or when $x=y = 1$ . Then $x+y = 2 \not < 2$ . That is there is no $(x,y)$ that satisfies the equality condition, therefore the minimum value of $xy + \dfrac 1{xy}$ is not 2. The statement is false .

AM-GM Inequality!

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