What is the minimum value of f(x,y) under a symmetric constraint?

[anemone]

Let $m,\,n,\,p,\,q$ be the positive real numbers such that $\dfrac{1}{1+m^4}+\dfrac{1}{1+n^4}+\dfrac{1}{1+p^4}+\dfrac{1}{1+q^4}=1$, find the minimum of the product of $mnpq$.

 

[MarkFL]

Here is my solution:

Spoiler

Because of cyclic symmetry*, we know the extremum will occur for:

$$m=n=p=q$$

which gives us:

$$\frac{4}{1+m^4}=1$$

$$m^4+1=4$$

$$m^4=3$$

Hence, the product $mnpq=3$.

* When any two of the set of variables in an optimization problem with constraint can be switched without changing either the objective function or the constraint, then we have cyclic symmetry. You will always find in such a case that the method of Lagrange multipliers will imply that the critical point will occur when all of the variables are equal to one another. So, a nice shortcut to such a problem where you see there is cyclic symmetry is to let all of the variables be equal and solve for the value using the constraint. This alleviates the need to compute the partials.

 

[anemone]

Thanks MarkFL for participating and thanks for your solution using the nice shortcut as well.

Here is a solution that used the trigonometric approach that I'd like to share with MHB:

Spoiler

If we let $m^2=\tan M$, $n^2=\tan N$, $p^2=\tan P$, and $q^2=\tan Q$, where $M,\,N,\,P,\,Q\in \left(0,\,\dfrac{\pi}{2} \right)$, then the original equality becomes $\cos^2 M+\cos^2 N+\cos^2 P+\cos^2 Q=1$.

By AM-GM, we have $\sin^2 M=1-\cos^2 M=\cos^2 N+\cos^2 P+\cos^2 Q\ge3(\cos N \cos P \cos Q)^{\dfrac{2}{3}}$.

Similarly we have

$\sin^2 N ge3(\cos P \cos Q \cos M)^{\dfrac{2}{3}}$,

$\sin^2 P ge3(\cos Q \cos M \cos N)^{\dfrac{2}{3}}$,

$\sin^2 Q ge3(\cos M \cos N \cos P)^{\dfrac{2}{3}}$,

Multiplying these four inequalities yields the result.

 

[Opalg]

Here is my solution:

Spoiler

Because of cyclic symmetry*, we know the extremum will occur for:

$$m=n=p=q$$

which gives us:

$$\frac{4}{1+m^4}=1$$

$$m^4+1=4$$

$$m^4=3$$

Hence, the product $mnpq=3$.

* When any two of the set of variables in an optimization problem with constraint can be switched without changing either the objective function or the constraint, then we have cyclic symmetry. You will always find in such a case that the method of Lagrange multipliers will imply that the critical point will occur when all of the variables are equal to one another. So, a nice shortcut to such a problem where you see there is cyclic symmetry is to let all of the variables be equal and solve for the value using the constraint. This alleviates the need to compute the partials.

That answer is correct, of course. But this method of "cyclic symmetry" needs to be used with caution. To take a simple example, what is the minimum value of $f(x,y) = \bigl(x^2 + (y-1)^2\bigr)\bigl(y^2 + (x-1)^2\bigr)$ subject to the constraint $x+y=1$? The function and the constraint are symmetric in $x$ and $y$ (and clearly $f(x,y)\to\infty$ as $|x|$ and $|y|$ go to infinity), so the method indicates that the minimum will occur when $x=y=\frac12$, with $f\bigl(\frac12, \frac12\bigr) = \frac14.$ But in fact $f(1, 0) = f(0, 1) = 0.$ The explanation in that case is that there is indeed an extreme point at $\bigl(\frac12, \frac12\bigr)$, but it is a local maximum. There are also two "unexpected" extreme points, $(1,0)$ and $(0,1)$, away from the symmetric point, where the minimum occurs.