Find smallest $a$, that satisfy $5ax^2+3\geq2$ $\space \forall x\in \mathbb{R}$ I need to find the smallest $a$ that satisfy the inequality $\forall x\in \mathbb{R}$: $$5ax^2+3\geq2$$
It is a very simple problem. I can easily see it would give problems if $a<0$. And it seems obvious that $a\geq0$ satisfy the inequality $\forall x\in \mathbb{R}$. So I should set $a=0$.
But how do I solve/argue for this more rigidly? What if it wasn't a simple polynomial. Like what method should I use to find this a, if it isn't obvious by just looking at the equation?
 A: $$5ax^2 \ge -1$$
If $x=0$, this clearly holds.
If $x \ne 0$, we need $a \ge -\frac1{5x^2}$.
Hence we should pick $a$ to be the supremum of $\{-\frac1{5x^2}: x \in \mathbb{R} \setminus \{0\} \}$
$-\frac1{5x^2}<0$ and in fact, we can make it arbitrary close to $0$ since $-\lim_{x \to \infty}\frac1{5x^2}=0$.
Hence we should pick $a=0$.

If $a \ge 0$, we have $5ax^2 \ge 0 > -1$.
If $a<0$, we have $\lim_{x \to \infty}5ax^2 = -\infty < -1.$
A: Case $1$: $a\geq 0$. For the polynomial $5ax^2+1$ to be greater than or equal to $0$, it needs to have a non-positive determinant, that is, $$-4 \cdot 5a \leq 0 \Longrightarrow a\leq 0\Longrightarrow a=0$$
Case $2$: $a\leq 0$. Let $a=-b$ where $b\geq0$. Again, the polynomial needs to have a non-positive determinant, that is,
$$-4\cdot 5a=-20(-b)=20b\leq0\Longrightarrow b\leq 0 \Longrightarrow b=0\Longrightarrow a=0 $$
In both cases, we were forced to choose $a=0$ for all choices of $x$.
If $a<0$, then there exists an $x$ such that $5ax^2+1<0$, that is, any $x$ on the interval  $(\frac{\sqrt{-a}}{5a},-\frac{\sqrt{-a}}{5a})$ will break the inequality.
Any element in the set $\{a\in\mathbb{R}:a\geq0 \}$ will satisfy the inequality, and the smallest such element is $a=0$.
A: Start with $5ax^2\ge -1$.  When $a\ge 0$, $5ax^2 \ge 0$, since $5x^2\ge 0$ and the product of two non-negative numbers is non-negative.
A: If $a<0$ then $$\{5ax^2+3: x\in \Bbb R\}=\{5ay+3: y\ge 0\}=$$ $$=\{a(5y)+3:y\ge 0\}=$$ $$=\{a(z)+3: z\ge 0\}=$$ $$=\{b+3:b\le 0\}=$$ $$=(-\infty,3]\supset \{0\}.$$ So with $D=\{a:\forall x\in \Bbb R\,(5ax^2+3\ge 2)\}$ we have $D\subseteq [0,\infty).$ And $0\in D.$
Since $0\in D\subseteq [0,\infty)$, we have  $\min D=0.$
A: Suppose you have a parametric family of functions $f_a$ defined on the real line, where $a$ is the parameter. The following:
$$ f_a(x)\geq 0 \qquad \forall x \in \mathbb{R}\tag{1}$$
is equivalent to:
$$ \inf_{t \in \Bbb{ R}} f_a(t) \geq 0 \tag{2}$$
Therefore finding the values of $a$ for which $(1)$ holds leads to the equivalent problem of finding the values of $a$ for which $(2)$ holds.
In your case:
$$5ax^2+3\geq2 \Leftrightarrow 5ax^2+1\geq0$$
and setting $f_a(x):=5ax^2+1$ we have:
$$\inf_{t \in \Bbb{ R}} f_a(t) = \begin{cases} -\infty & a<0 \\ 1 & a\geq 0\end{cases}$$
This shows $(1)$ holds if and only if $a\geq 0$. Once you know for which values of $a$ inequality $(1)$ holds, take the infimum of these values and call it $\bar{a}$ ($\bar{a} = \inf [0,+\infty) = 0$ in your case). The last step is checking whether $a=\bar{a}$ is finite and satisfies $(1)$. If that is the case, then $\bar{a}$ is the smallest value of $a$ satisfying $(1)$. Otherwise, such a smallest value does not exist.
EDIT:
Checking explicitly if $\inf_{t \in \Bbb{ R}} f_a(t) \geq 0$ can be a tough problem in general. For the easiest cases a standard study of functions is sufficient.
