For all $x \in \Bbb R$ there exists a $y \in (-\infty, 1)$ such that $3yx^2+y-x=0$

Question

I have this question:

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

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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.

Answer 1

The universe is the set of real numbers.

$\forall x\exists y<1:3x^2y+y-x=0$

$\iff\forall x\exists y<1:3x^2y+y=x$

$\iff\forall x\exists y<1:y(3x^2+1)=x$

$\iff\forall x\exists y<1:y=\frac{x}{3x^2+1}$, because we know that $3x^2+1\not=0$

$\iff\forall x\exists y(y<1\land y=\frac{x}{3x^2+1})$

$\iff\forall x\exists y(y=\frac{x}{3x^2+1}\land y<1)$

$\iff\forall x\exists y(y=\frac{x}{3x^2+1}\land \frac{x}{3x^2+1}<1)$

$\iff\forall x(\exists y(y=\frac{x}{3x^2+1})\land \frac{x}{3x^2+1}<1)$

$\iff\forall x\frac{x}{3x^2+1}<1$

$\iff\forall x:x<3x^2+1$, because we know that $3x^2+1>0$

$\iff\forall x:3x^2-x+1>0$

$\iff\forall x:x^2-\frac{1}{3}x+\frac{1}{3}>0$

$\iff\forall x:(x-\frac{1}{6})^2-(\frac{1}{6})^2+\frac{1}{3}>0$

$\iff\forall x:(x-\frac{1}{6})^2-\frac{1}{36}+\frac{1}{3}>0$

$\iff\forall x:(x-\frac{1}{6})^2>\frac{-11}{36}$, which is true, because the square of every real number is positive.

Answer 2

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.

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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.