Prove that a multivariable polynomial has no root

Question

I want to find when the gradient of the following function vanishes $$ f(x,y) = \ln(1+x^2 + 2y^2 ) + x^2 y^2 + y.$$ Nevertheless I have trouble to solve one part. I have to find when this polynomial $$ 2x + 2y^2 + 2 x^2 y^2 + 4y^4 $$ is equal to $0$. I tried to use the method of completing the squares (of Gauss I assume) but in vain. I have the intuition that it never goes through zero, but I have no proof.

Answer 1

Taking

$$ f(x,y) = \ln(1+x^2 + 2y^2 ) + x^2 y^2 + y $$

we have

$$ \nabla f = \left\{ \begin{array}{rcl} 2 x \left(y^2+\frac{1}{x^2+2 y^2+1}\right) & = & 0\\ 2 y x^2+\frac{4 y}{x^2+2 y^2+1}+1 & = & 0 \end{array} \right. $$

and real solutions given by

$$ x = 0,\; y = \frac{1}{2}(-2\pm \sqrt2) $$

Answer 2

I think that your computation of the gradient of $f$ is wrong somewhere. Note that $$f_x(x,y)=\frac{2x}{1+x^2+2y^2}+2xy^2=2x\cdot\underbrace{\left(\frac{1}{1+x^2+2y^2}+y^2\right)}_{>0}=0$$ if and only if $x=0$. Now evaluate $f_y$ and solve $f_y(0,y)=0$ with respect to $y$.

Can you take it from here? What are the stationary points of $f$?