Prove that $\frac{x^2y}{1+x^4+y^2}$ has no global minimum

Question

I am not sure how to approach this problem. The usual methods do not work to find a minimum.

I can see that, but how to show that there must not exist a minimum?

Answer 1

Suppose that it has a global minimum at $(a,b)$. Then the map $y\mapsto f(a,y)$ has a minimum at $b$. But then the standard Calculus methods show that $b=-\sqrt{1+a^4}$ and that tharefore the minimum is $-\frac{a^2}{2\sqrt{1+a^4}}$. This cannot be, because$$\lim_{a\to\pm\infty}-\frac{a^2}{2\sqrt{1+a^4}}=-\frac12,$$but the limit is never attained.

Answer 2

Since $f(0,y) = 0$, we see that $y$ would have to be negative for a global minimum, and $x$ would have to be nonzero.

However, on those conditions:

$$\begin{array}{rcl} f(2x,4y) &=& \dfrac{(2x)^2(4y)}{1+(2x)^4+(4y)^2} \\ &=& \dfrac{16x^2y}{1+16x^4+16y^2} \\ &=& \dfrac{x^2y}{0.0625+x^4+y^2} \\ &<& \dfrac{x^2y}{1+x^4+y^2} \\ &=& f(x,y) \end{array}$$

Answer 3

One heuristic is to look at values where the terms in the sum are equal, that is $x^4=y^2$. Since we are interested in negative values this leads to $y=-x^2$.

You can easily check that the limit for large $x$ is now $-1/2$.

It is now sufficient to show that the expression is always larger than $-1/2$, but this can be done by multiplying the desired inequality by the denominator and recognize three terms of a complete square.

Answer 4

Since $(x^2-|y|)^2 = x^4+y^2 - 2 x^2|y| \ge 0$, we have $x^4+y^2 \ge 2 x^2 |y|$.

Hence ${x^2|y| \over 1+x^4+y^2} \le {x^4+y^2 \over 1+x^2 + y}<{1 \over 2}$.

If we let $x(t) = \sqrt{|t|}, y(t) = -t$, we have $\lim_{t \to\infty } {-t^2 \over 1+ 2 t^2} = -{1\over 2}$.

In particular, $\inf_{x,y} {x^2 y \over 1+x^4+y^2} = - {1 \over 2}$, but the value of $-{1 \over 2}$ is never attained.

Answer 5

For all $x,y$, we have

$$\begin{align} x^4+2x^2y+y^2=(x^2+y)^2\ge0\gt-1 &\implies2x^2y\gt-1-x^4-y^2\\ &\implies{x^2y\over1+x^4+y^2}\gt-{1\over2} \end{align}$$

but, setting $y=-x^2$, we have

$${-x^4\over1+2x^4}\to-{1\over2}$$

as $x\to\infty$, so $-1/2$ is an infimum that is never attained.

Answer 6

$$f(x,y) = \frac{x^2y}{1+x^4+y^2}$$

$$\frac{\partial f}{\partial x} = \frac{2xy(1+x^4+y^2) - 4x^5y}{(1+x^4+y^2)^2}$$

$$\frac{\partial f}{\partial y} = \frac{x^2(1+x^4+y^2) - 2x^2y^2}{(1+x^4+y^2)^2}$$

$$\frac{\partial f}{\partial x} = 0 \implies 1+x^4+y^2 = 2x^4 \lor x=0 \lor y=0$$

$$\frac{\partial f}{\partial y} = 0 \implies 1+x^4+y^2 = 2y^2 \lor x=0$$

One check that when $x=0$ or $y=0$, the function is zero, so it is not the global minimum.

Otherwise, on solving the equations we have $y=x^2$, and $1+2x^4=2x^4 \implies 1 = 0$.