Bound for the roots of a polynomial in terms of coefficients

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

• Hi @Conrad! First up, thank you for the response. I am currently studying your sketch, and I seem to be stuck on step 2 because I can't prove the inequality. Is there an easier way to see it? Again, thank you! – chhro Mar 10 at 15:57
• Just checking in to say that I now fully understand all steps except 2 which I will think about more. Highly appreciate the help! – chhro Mar 10 at 17:22
• Show 2 with $b=c_1$ and all others $a$ - this is straightforward by passing the $b$ term and subtracting and the induction or just step by step subtracting from the highest term and then show that that's best you can do as position b goes – Conrad Mar 10 at 17:44
• Thanks for the clarification @Conrad! That settles it. – chhro Mar 11 at 16:18