Find absolute minimum of $f(x,y,z,w) = x^2 y + y^2 z + z^2 w + w^2 x$ on $S = \{x>0,y>0,z>0,w>0, xyzw=1\} \subset \mathbb{R}^4$

Question

I tried to solve it using the Lagrange multiplier method, but I failed since the system of equations involving partial derivatives was pretty complex.

Any hints will be appreciated.

Answer 1

You can use the inequality of arithmetic and geometric means to get

$$\frac{x^2 y + y^2 z + z^2 w + w^2 x}{4} \ge \sqrt[4]{(xyzw)^3} = 1 \implies x^2 y + y^2 z + z^2 w + w^2 x \ge 4 \tag{1}\label{eq1A}$$

The right side lower limit of $4$ is reached with $x = y = w = z = 1$.

Answer 2

The system of Lagrange equations looks complicated, but it possesses an interesting property:

$$ 2xy \ + \ w^2 \ = \ \lambda · yzw \ \ , \ \ x^2 \ + \ 2yz \ = \ \lambda · xzw \ \ , $$ $$ y^2 \ + \ 2zw \ = \ \lambda · xyw \ \ , \ \ z^2 \ + \ 2wx \ = \ \lambda · xyz \ \ . $$

The amount of symmetry among these equations is already suspicious. We can make these easier to work with by applying the constraint equation $ \ xyzw = 1 \ $ to write

$$ 2xy \ + \ w^2 \ = \ \lambda · \frac{1}{x} \ \ , \ \ x^2 \ + \ 2yz \ = \ \lambda · \frac{1}{y} \ \ , $$ $$ y^2 \ + \ 2zw \ = \ \lambda · \frac{1}{z} \ \ , \ \ z^2 \ + \ 2wx \ = \ \lambda · \frac{1}{w} $$

$$ \Rightarrow \ \ \lambda \ \ = \ \ 2x^2y \ + \ xw^2 \ \ = \ \ x^2y \ + \ 2y^2z \ \ = \ \ y^2z \ + \ 2z^2w \ \ = \ \ wz^2 \ + \ 2w^2x \ \ . $$

If we combine these equations pairwise, we produce

$$ x^2y \ = \ 2y^2z - xw^2 \ \ , \ \ y^2z \ = \ 2z^2w - x^2y \ \ , \ \ z^2w \ = \ 2w^2x - y^2z \ \ . \ \ w^2x \ = \ 2x^2y - wz^2 \ \ . $$

It is possible to put pairs of these equations together to find the correspondence between the variables $ \ x \rightarrow y \ , \ y \rightarrow z \ , \ z \rightarrow w \ , \ w \rightarrow \ x \ \ . $ This interchangeability among the variables indicates that $ \ x = y = z = w \ \ , $ and using this result in the constraint equation tells us that all four are equal to $ \ 1 \ $ . Hence, we have an absolute minimum for the function of

$$ f(1,1,1,1) \ = \ 4 \ · \ 1^2 · 1 \ = \ 4 \ \ , $$

since there is no upper limit to the values of the variables.

[As a check, a small perturbation away from the minimum by setting, say, $ \ x = 1 + \epsilon \ , \ y = \frac{1}{1+\epsilon} \ \ , $ produces

$$ f \left(1+\epsilon,\frac{1}{1+\epsilon},1,1 \right) \ = \ \frac{(1+\epsilon)^2}{1+\epsilon} \ + \ \frac{1}{(1+\epsilon)^2} · 1 \ + \ 1 · 1 \ + \ 1^2 · (1+\epsilon) $$ $$ = \ \ (1+\epsilon) \ + \ ( 1 - 2\epsilon + 3\epsilon^2 - 4\epsilon^3 + 5\epsilon^4 + \ldots) \ + \ 1 \ + \ (1+\epsilon) $$ $$ = \ \ 4 + 3\epsilon^2 - 4\epsilon^3 + 5\epsilon^4 + \ldots \ \ . \ ] $$