Task Let $p:=(0,0)$ and $f: \mathbb{R}^{2} \longrightarrow \mathbb{R}$ be the function $$f(x, y):=y^{2}-3 x^{2} y+2 x^{4}$$ Show:
a) For every $a=\left(a_{1}, a_{2}\right) \in \mathbb{R}^{2} \backslash\{(0,0)\}$, the function $g_{a}: \mathbb{R} \longrightarrow \mathbb{R}$ $$g_{a}(t):=f(p+t a)$$ has a strict local minimum at $t=0$.
b) $f$ has no local minimum at $p$. Hint: Consider $f$ along the curve $\gamma: t \mapsto\left(t, \sqrt{2} t^{2}\right)$ passing through $p$.
My solution: a) We have $$g_{a}(t)=f(p+ta)=f((0,0)+t\cdot(a_1,a_2))=f(ta_1,ta_2)=t^2a_2^2-3t^3a_1^2a_2+2t^4a_1^4$$ $$g_a'(t)=2ta_2^2-9t^2a_1^2a_2+8t^3a_1^4$$
Necessary condition:
$g_a'(t)=0$, $0=2ta_2^2-9t^2a_1^2a_2+8t^3a_1^4$, $0=t(2a_2^2-9ta_1^2a_2+8t^2a_1^4)$
Case 1
$t_1=0$
Case 2
$$\begin{aligned} 2a_2^2-9ta_1^2a_2+8t^2a_1^4 &= \frac{2a_2^2}{8a_1^4}-\frac{9a_1^2a_2}{8a_1^4}t+t^2 \\ &= \frac{a_2^2}{4a_1^4}-\frac{9a_2}{8a_1^2}t+t^2, \\ t_{2,3} &= \frac{9a_2}{16a_1^2}\pm\sqrt{\frac{81a_2^2}{256a_1^4}-\frac{64a_2^2}{256a_1^4}} \\ &= \frac{9a_2}{16a_1^2}\pm\frac{\sqrt{17}a_2}{16a_1^2}, \\ t_2 &= \frac{a_2(9+\sqrt{17})}{16a_1^2}, \\ t_3 &= \frac{a_2(9-\sqrt{17})}{16a_1^2} \end{aligned}$$
Sufficient condition $$g_a''(t)=2a_2^2-18a_1^2a_2t+24t^2a_1^4$$ Then, $$g_a''(t_1)=2a_2^2>0$$ and $$\begin{aligned}0<t^2a_2^2-3t^3a_1^2a_2+2t^4a_1^4,\\ 0<t^2(ta_1^2-a_2)(2ta_1^2-a_2)\end{aligned}(*)$$ $$\Rightarrow \text{True statement}$$ $$\Rightarrow t_1=0 \text{ is a strict local minimum}$$
Problems:
I'm struggeling with my approach of (*). And for b) I don't really have an idea but I substituted $(t,\sqrt{2}t^2)$ into $f$ and get $f(t,\sqrt{2}t^2) = t^4(\sqrt{16}-\sqrt{18})< 0$.
Thanks so much for your help!
You want to prove that $g_a =f(p + ta)$ as a local minimum for $t = 0$
Once you know that $g_a'(t)$ cancels out at $t = 0$ there is no point in wasting your time trying to find the other roots of $g_a'$. Indeed you are only interested in proving that there is a minimum at $t = 0. Solving
$$2a_2^2-9ta_1^2a_2+8t^2a_1^4 = 0$$ is just a waste of time. Now that you know that $t = 0$ cancels out $g'_a$ consider the sign of the second derivative at $t = 0$ only ! Your calculations show that the second derivative is positive at $t = 0$ so we have a local minimum for $g_a$ at $t = 0.$
This solves $a)$
For $b)$ use the hint and consider $$f(t,\sqrt 2 t^2) = 2t^4 -3t^2 \sqrt 2 t^2 + 2 t^4 = t^4(4-3 \sqrt 2)$$ which is $\leq 0$ for all values of $t$.
It follows that the function $f$ has no local minima at $p$.