Given that a, b, and c are real numbers, and the equation $|az^3+bz^2+c|$$\le$ 1 holds for all complex numbers $z$ that satisfy $|z|\le$ 1, find the maximum value of $ 7 (a+c) -5b$.
I used the maximum modulus principle and substitution, and found that it is equivalent to $f(x,a,b,c)=8acx^3+4bx^2+2a(b-3c)x+a^2+(b-c)^2$
Let f obtain its maximum value at $\hat{x}$ , and the Lagrange multiplier is
$L(a,b,c,\lambda)=7(a+c)-5b- \lambda(f(\hat{x},a,b,c)-1)$
then use software to solve,getting the solution $(a,b,c,\lambda,\hat{x})=(\frac{2}{\sqrt{13}},-\frac{1}{\sqrt{13}},\frac{1}{\sqrt{13}},\sqrt{13},-\frac{1}{2})$ At that time,$7(a+c)-5b=2\sqrt{13}$
this solution is too complicated，I want a nice solution
Hope to see some useful help here. Thank you
Remarks: Once we motivate the solution by some way, we can write a simple solution, even though the motivation behind the solution is complicated.
First, we prove that $7(a + c) - 5b \le 2\sqrt{13}$.
Letting $z = -\frac12 + \frac{\sqrt 3}{2}\mathrm{i}$, from $|az^3 + bz^2 + c| \le 1$, we have $$a^2 + b^2 + c^2 - ab - bc + 2ca \le 1. \tag{1}$$
Using (1), we can prove that $7(a + c) - 5b \le 2\sqrt{13}$. For example, we have $$52(a^2 + b^2 + c^2 - ab - bc + 2ca) - (7a + 7c - 5b)^2 = 3(a + 3b + c)^2\ge 0.$$
Second, we prove that $7(a + c) - 5b = 2\sqrt{13}$ is feasible.
Let $a = \frac{2}{\sqrt{13}}, b = -\frac{1}{\sqrt{13}}, c = \frac{1}{\sqrt{13}}$.
By the maximum modulus principle, it suffices to prove that $|az^3 + bz^2 + c| \le 1$ for all $|z| = 1$. Let $z = \cos x + \mathrm{i} \sin x$. After some manipulations, $|az^3 + bz^2 + c|^2 \le 1$ is written as $\frac{5 - 4\cos x}{13}(1 + 2\cos x)^2 \ge 0$ which is true.
(Note: If we don't use the maximum modulus principle, we can still prove it.)
We are done.