I am trying to prove an Exercise in Bhatia's Matrix Analysis, and I'm unsure how to approach the problem.
Let $f(z)=z^n+a_1z^{n-1}+\cdots+a_n=(z-\lambda_1)\cdots(z-\lambda_n)$ be a given monic polynomial. Let $\mu_1,\ldots,\mu_n$ be the numbers $|a_k|^{1/k}$ for $k=1,\ldots,n$, rearranged in decreasing order. Show that the roots satisfy $$|\lambda_i|\leq\mu_1+\mu_2$$ for all $i=1,\ldots,n$.
Any help is greatly appreciated.
I will sketch an answer and be happy to add details if needed:
1: if all $a_k$ are zero, roots are zero, problem clear, so assume at least one of them is non-zero; also we can assume $n \geq 2$ as the problem is clear by inspection
2: let $0 \leq a \leq b$ real numbers, and $n \geq 2$; we can prove an inequality: $c_n^n+c_{n-1}^{n-1}(a+b)+c_{n-2}^{n-2}(a+b)^2+..c_1(a+b)^{n-1} \leq (a+b)^n$, where $c_k$ are all $a$ except one that is $b$; the proof for example works by showing that LHS increases if the place of the one $b$ is lower (as index) and then considering the case $c_1=b$ which is fairly straightforward.
3: consider the Cauchy polynomial associated to $f$, namely $g(z) = f(z)=z^n- |a_1|z^{n-1}-|a_2|z^{n-2}-\cdots-|a_n|$
It is easy to see that (assuming not all $a$ zero) $g$ has an unique positive roof $c(f)$ (divide by $z^n$ and show the coefficient part is decreasing in $r>0$) and then using the triangle inequality $|\lambda_i|\leq c(f)$ when $\lambda_i$ is any of the roots of $f$; in particular if $R>0$ satisfies $g(R) \geq 0$, then $|\lambda_i|\leq R$ for any root of $f$
4: using the given coefficient inequality with $a=\mu_2, b=\mu_1$, and the definition of $\mu_1, \mu_2$ it follows that $|a_k| \leq a^k$ for all but one $k$, where $|a_k| =b^k$
5: using the inequality at point 2: it then follows $g(a+b) \geq 0$, so using point 3: we are done
