Let $(K,|\cdot|)$ be a complete discretely valued field, where $|\cdot|$ is a non-archimedean absolute value (or multiplicative valuation). Fix an algebraic closure $\bar{K}$ of $K$. We also denote by $|\cdot|$ the unique absolute value on $\bar{K}$ extending the one on $K$. On the polynomial ring $K[T]$ we define the Gauss norm: for $P(t)=a_n t^n+a_{n-1} t^{n-1}+\ldots+a_0 \in K[t]$, let $||P(t)||:=\max\{|a_0|,|a_1|,\ldots |a_n|\}$.
Consider a monic polynomial $f\in K[t]$ and let $\alpha\in \bar{K}$ be a root of $f$. Show that $|\alpha|\le ||f||$
Now it is clear that if $\displaystyle f(t)=t^n+\sum^{n-1}_{i=1}c_it^i$, then $$|\alpha^n|=\displaystyle |\sum^{n-1}_{i=1}c_i\alpha^i|\le \max\{|c_i \alpha^i|\}$$ Hence for some $i_o\in \{1,2,...,n-1\}$, we have that $|c_{i_0}|^{\frac{1}{n-i_0 }}\ge |\alpha|$, hence if $|\alpha|\ge 1$, we're done.
However, for the case $|\alpha|< 1$, I am not able to prove this. Also, I am in fact getting the opposite inequality when say $f$ is the minimal monic irreducible polynomial of $\alpha$, because say $f=(x-\alpha)(x-\sigma_2(\alpha))...(x-\sigma_n(\alpha))$, then $|\sigma_i(\alpha)|=|\alpha|$ and $$|c_{n-j}|=|\sum \sigma_{\tau(1)}(\alpha)\sigma_{\tau(2)}(\alpha)...\sigma_{\tau(n)}(\alpha)|\le \max| \sigma_{\tau(1)}(\alpha)\sigma_{\tau(2)}(\alpha)...\sigma_{\tau(n)}(\alpha)|=|\alpha|^j<|\alpha|$$ if $j>1$, so the only chance for the inequality to remain true is if $|\alpha|=|c_{n-1}|$, and I don't know what to do from here.