# Show that for a fixed $t\in\mathbb{R}$, $F(y)=y^4+ty^2+t^2y$ achieves its global minimum at a single point

For any fix $$t\in\mathbb{R}$$, show that the function $$F_{t}(y)=y^4+ty^2+t^2y$$ achieves its global minimum at a single point $$y_{0}(t)$$. I have made some partial attempts to solve this question, but I am stuck at one of the steps and would like to ask for hints.

$$\textbf{My attempt}$$: Fix any $$t\in\mathbb{R}$$. Since $$\lim_{y\rightarrow\infty}F_{t}(y)=\lim_{y\rightarrow -\infty}F_{t}(y)=\infty$$, $$F_t(y)$$ achieves its global minimum at a point $$y_{0}$$. We now show that there can only be one values of $$y$$ where $$F_{t}(y)$$ achieves its global minimum.

Case 1: If $$t=0$$ then it is quite obvious.

Case 2: Assume $$t>0$$, then $$\frac{dF_{t}(y)}{dy}=4y^3+2ty+t^2$$, and $$\frac{d^2F_{t}(y)}{dy^2}=12y^2+2t$$. Since the second derivative of $$F_{t}(y)$$ is always positive, the first derivative of $$F_{t}(y)$$ is always strictly increasing, and hence must be injective. Furthermore, $$\lim_{y\rightarrow -\infty}\frac{dF_{t}(y)}{dy}=-\infty$$, and $$\lim_{y\rightarrow\infty}\frac{dF_{t}(y)}{dy}=\infty$$. By continuity and previous argument, the first derivative of $$F_{t}(y)$$ has exactly one zero. Therefore, $$F_{t}(y)$$ achieves its global minimum only at a single point $$y_{0}(t)$$.

Case 3: t<0. I am stuck at this step. A hint would be useful.

$$((x-a)(x-b))^2+c.$$
As there is no cubic term, $$a=-b$$ and the equation is in fact
$$(x^2-a)^2+c=x^4-2ax^2+a^2+c.$$
As there is no linear term, we must have $$t=0$$, which also imposes $$a=0$$.