Minimum of A

Algebra Level 5

$\large A = \dfrac{x-y}{x^4+y^4+6}$

Let $x$ and $y$ be real numbers. Find the minimum value of $A$ .

The answer is -0.25.

3 solutions

Sam Bealing
Apr 15, 2016

Clearly a minimum will occur for negative $x$ and positive $y$ (otherwise replace $x$ with $-x$ and $y$ with $-y$ and you have a smaller value for $A$ .

Substitute $z=-x$ so $z$ and $y$ are therefore positive.

The expression becomes $A=-(\frac{z+y}{z^4+y^4+6})$

We have that $z^4+y^4+6=(z^4+1+1+1)+(y^4+1+1+1) \geq 4z+4y$ by AM-GM on the two brackets as they are both positive.

This gives: $A=-(\frac{z+y}{z^4+y^4+6}) \geq -(\frac{z+y}{4z+4y}) =\boxed{-0.25}$ with the minimum occuring when $z=y=1$ giving $x=-1,y=1$

Moderator note:

Nice application of AM-GM to this problem, due to the nice setup of the expression.

That is a really beautiful solution ! Thank you so much ^^

Khoa Đăng - 5 years, 2 months ago

Otto Bretscher
Apr 15, 2016

Substitute $x=u-v,y=u+v$ to find $A=-\frac{2v}{2v^4+6+2u^4+12u^2v^2} \geq -\frac{v}{v^4+3}$ for $v>0$ . By AM-GM we have $v^4+1+1+1\geq 4v$ so that $-\frac{v}{v^4+3}\geq -\boxed{0.25}$ with equality when $v=1$ .

How do you know we must start with $x=u-v, y=u+v$ ?

Pi Han Goh - 5 years, 1 month ago

I'm thinking Linear Algebra (as I always do, in the back of my mind): I'm doing a change of coordinates, a rotation-scaling, to make $x-y$ into a single variable.

Otto Bretscher - 5 years, 1 month ago

So what are the steps that lead you to do these substitutions? Consider that I'm not familiar with "change of coordinates + rotation-scaling + linear algebra".

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – It's quite a "natural" substitution: $u$ is the average of $x$ and $y$ , and $\pm v$ is their deviation from the average.

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Where can I learn this natural substitution? From this book ?

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – From any good linear algebra text, including mine ;)

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – What keyword should I be looking for? Kernel? Spectral? Matrix? Otto? Comrade?

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – Keyword to find what now?

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – This natural substitution technique.

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – It's not an "official" technique, but if you understand change of coordinates well, it becomes natural ;) In my text this topic is discussed in Section 3.4, "Coordinates"

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Well, it's certainly inconvenient for me to buy that book just to read that one section. Do you have a simpler example that illustrates/uses this natural technique as well?

Pi Han Goh - 5 years, 1 month ago

Manish Bhat
Apr 20, 2016

There is a method which is slightly more tedious than the beautiful solutions listed, but probably easier to see. Set partial derivative of A with respect to x and y=0. From this one gets 2 equations of form f(x,y)=0 which can be solved to give two possible values (1,-1) and (-1,1). Minima is observed for (-1,1)