Min-max theorem If $f$ is a polynomial. Prove that there is a $y\in \mathbb{R}$ which $|f(y)|\le |f(x)|,$ for every $x\in \mathbb{R}$:
I said since $f$ is a polynomial it's continuous in $\mathbb{R}$ so we can use the theorem of $a=\min, \max=b$ value, so $f(x_a)\le f(y)\le f(x_b)$ assuming that $f(x_a) = -f(x_b)= -f(x)$ then we got $-f(x)\le f(y)\le f(x)  \le   |f(y)|\le f(x)\le |f(x)|$, proved
Is this correct?
 A: You need to work a bit harder. If $f$ is a constant polynomial, you can pick any $y$. So assume $f$ is not a constant. Then $\lim_{x\to\pm\infty}|f(x)|=\infty$. So, fix any $c\in\mathbb{R}$. By the limit statement before, you can find $a, b\in\mathbb{R}$ so that $|f(x)|>|f(c)|$ whenever $x<a$ or $x>b$. Now, can you see how to go the rest of the way?
A: Suppose $f$ has a root. Then this root is your minimum.
Suppose now $f$ has no root, and wlog we assume $f>0$. [1]. Then $f$ has a derivative which must (reason to the degree) have a root. [2] But one of it's roots is therefore a global minimum. This states the claim.
Edit: [1]: This means the leading coefficient of the polynomial has to be positive.
[2]: Also, the leading coefficient has to be non-negative, which states that at least one of the roots leads to a minimum.
A: Let $f(x)$ be a polynomial of degree $n$. Since $|f(x)|\geq0$ for all $x\in\Bbb R$, clearly if it has a root, then we can take this for $y$ and $|f(y)|=0\leq|f(x)|$ for all $x\in\Bbb R$. Otherwise, as Harald Hanche-Olsen points out, $$\lim_{x\to\pm\infty}|f(x)|=\infty.$$ Since a polynomial has at most $n-1$ turning points, we can find $a,b\in\Bbb R$ such that $|f(x)|$ has no turning points for $x<a$ and $x>b$. This means the minimum of $|f(x)|$ is contained in $[a,b]$ (otherwise it will have a turning point, which is a contradiction). Now we can apply the Extreme Value Theorem to conclude $|f|$ has a minimum, i.e., there exists a $y\in[a,b]\subseteq\Bbb R$ such that $|f(y)|\leq|f(x)|$ for all $x\in\Bbb R$.
