Inspired by Shivam Jadhav

Algebra Level 3

Over all ordered pairs of real numbers, what is the minimum value of

( x + y ) 2 + ( x y − 1 ) 2 ? (x+y)^2 + (xy-1) ^2 ?

Inspiration

0 Minimum is never achieved 4 1

Calvin Lin by Calvin Lin

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2 solutions

Alan Yan
Sep 7, 2015

( x + y ) 2 + ( x y − 1 ) 2 = x 2 + 2 x y + y 2 + x 2 y 2 − 2 x y + 1 = (x+y)^2 + (xy -1 )^2 = x^2 + 2xy + y^2 + x^2y^2 - 2xy + 1 =

= x 2 y 2 + x 2 + y 2 + 1 = ( x 2 + 1 ) ( y 2 + 1 ) = x^2y^2+x^2+y^2+1 = (x^2+1)(y^2+1)

Both factors are positive and are minimal when x = y = 0 x = y = 0 .

Therefore the answer is 1 \boxed{1}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

You don't even have to factor; you can just say that x 2 y 2 ≥ 0 x^2 y^2 \ge 0 , x 2 ≥ 0 x^2 \ge 0 , and y 2 ≥ 0 y^2 \ge 0 .

Jon Haussmann - 5 years, 9 months ago

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And of course, keeping in the spirit of this question, you should verify that equality can be achieved in all of these inequalities.

Calvin Lin Staff - 5 years, 9 months ago

I factored to avoid having to deal with verifying the values.

In my way, I have rigorously made it very \textbf{very} clear what the minimum is.

But I acknowledge the fact that you don't have to factor.

Alan Yan - 5 years, 9 months ago
Calvin Lin Staff
Sep 7, 2015

[This is not a solution.]

You are intentionally misled to think that the minimum is 0, since we have the sum of 2 squares. However, equality cannot be achieved since we would require x + y = 0 x +y = 0 and x y − 1 = 0 xy - 1 = 0 , which gives us x 2 = − 1 x^2 = - 1 .

Instead, the minimum occurs when x = y = 0 x = y = 0 , which gives us the value of 1. How can we show this?

2 Helpful 0 Interesting 0 Brilliant 0 Confused

How we can find the minimum or the max?

Do you have any link for wiki so one with little knowledge grasp the concept ?

T h a n k s \color{grey}{Thanks}

Syed Baqir - 5 years, 9 months ago

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See Alan's solution and Jon's comment.

This is a very simple problem, so check out Trivial Inequality .

Calvin Lin Staff - 5 years, 9 months ago

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