<s>[INST] For all $x \in \Bbb R$ there exists a $y \in (-\infty, 1)$ such that $3yx^2+y-x=0$ I have this question:

Prove that for all real numbers $x$ there exist a number $y$ in the interval $(-\infty, 1)$ such that $$3x^2y+y-x=0$$


I proved that the range of the function
$$y = \frac{x}{3x^2 + 1}$$
is $(-1/2\sqrt{3},1/2\sqrt{3})$.
But the teacher said I should use another way.
I didn't find any.
 [/INST] Observe that, the given polynomial is exactly quadratic with respect to the variable $x$. You have:
$$
\begin{align}
&3x^2y+y-x=0\\
\iff &3yx^2-x+y=0\\
&\Delta_x=1-12y^2\ge 0\\
\iff &-\frac {1}{2\sqrt 3}\leq y\leq \frac {1}{2\sqrt 3}
\end{align}
$$
This result tells us, $\forall x\in\mathbb R$ if $3x^2+y-x=0$, then $-\frac {1}{2\sqrt 3}\leq y\leq \frac {1}{2\sqrt 3}$ which is logically equivalent to the original statement.
Finally, we see that
$$
\begin{align}
-&\frac {1}{2\sqrt 3}\leq y\leq \frac {1}{2\sqrt 3}\\
\implies &y\in\left(-\infty, 1\right).
\end{align}
$$
This completes the proof.
You can also get a stronger result here:

$\forall x \in \mathbb R$, there exists a $y \in \left[-\frac {1}{2\sqrt 3},\frac {1}{2\sqrt 3}\right]$, such that
$$3yx^2+y-x=0$$ holds.


Remember that, you can also reach the result in a shorter way:
We observe that, if $x≤0$ the statement is obviously trivial. Therefore, we analyze the case $x>0$. Then applying the AM-GM inequality, you have:
$$
\begin{align}
y=\frac {x}{3x^2+1}&<\frac{x}{2\sqrt 3x}\\
&=\frac {1}{2\sqrt 3}\\
&<1.\end{align}
$$
This also proves the original statement.
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