Consider the polynomial
$$ P(x) = x^{n} + a_{n-1}x^{n-1} + ... +a_{1} x + a_{0} $$
with only real roots and coefficients. If $R$ is the smallest bound such that $|x_{i}| \le R$ (it is the bound for the largest root), $P(x_{i})=0, i=0,1,...n-1$, then for every conjugate $\frac{1}{p} + \frac{1}{q} = 1, p,q >1$, prove that
$$ R < \left( 1 + A_{p}^{q} \right)^{1/q}, A_{p} = \left( \sum |a_{i}|^{p} \right)^{1/p}$$
Attempt
We have
$$ P(x_{0}) = P(x_{1}) = ... = P_(x_{n-1}) = 0 $$
sum all of them to get
$$ nx^{n} + a_{n-1} \sum x_{i}^{n-1} + ... + a_{1} \sum x_{i} + n a_{0} = 0$$
Let us start from easy case, $n=2$:
with only two roots $x_{0},x_{1}$ then we have $-x_{0} - x_{1} = a_{1}$, and $ a_{0} = x_{1} x_{0}$. We wish to prove
$$ R < \left[1 + \left( |x_{1}x_{0}|^{p} + |x_{1} + x_{0}|^{p} \right)^{q/p} \right]^{1/q} $$
... Clearly we must have $R = max(|x_{1}|, |x_{0}|)$, WLG letit be $|x_{0}|$. Let us work backwards:
$$ |x_{1}+a_{1}| + |\frac{a_{0}}{x_{1}}| < 2\left( 1 \right)^{1/p} \left[1 + \left( |x_{1}x_{0}|^{p} + |x_{1} + x_{0}|^{p} \right)^{q/p} \right]^{1/q} $$
$$ |x_{1}+a_{1}| + |\frac{a_{0}}{x_{1}}| < \left( (2^{q})^{p} + (2^{q})^{p} \right)^{1/p} \left[1 + \left( |x_{1}x_{0}|^{p} + |x_{1} + x_{0}|^{p} \right)^{q/p} \right]^{1/q} $$
How to continue..?
We have to show that $P(x) = 0$ implies $|x| < \left( 1 + A_{p}^{q} \right)^{1/q}$.
If $A_p = 0$ then $P(x) = x^n$ and the implication clearly holds.
If $A_p > 0$ and $|x| \le 1$ then the desired inequality holds as well.
Therefore it suffices to consider the case that $A_p > 0$, $P(x) = 0$ and $|x| > 1$. Using Hölder's inequality we get $$ |x|^n = | a_0 + a_1 x + \ldots + a_{n-1}x^{n-1}| \le \left( \sum_{k=0}^{n-1} |a_k|^p \right)^{1/p} \left( \sum_{k=0}^{n-1} |x^k|^{q} \right)^{1/q} = A_p \left( \sum_{k=0}^{n-1} |x|^{kq} \right)^{1/q} \\ \implies |x|^{nq} \le A_p^q \left( \sum_{k=0}^{n-1} |x|^{kq} \right) \, . $$ To simplify the notation we set $$ u = |x|^q \, , a = A_p^q \, . $$ Then $$ u^n \le a (1 + u + \ldots + u^{n-1}) \\ \implies 1 \le a \left( \frac 1u + \ldots + \frac{1}{u^{n}}\right) < a \sum_{k=1}^\infty \frac{1}{u^k} = \frac{a}{u-1} \\ \implies u-1 < a $$ or $u < 1+a$, and that is exactly the desired estimate $|x| < \left( 1 + A_{p}^{q} \right)^{1/q}$.
Remark: We haven't used the fact that $P$ has only real coefficients and roots, so this estimate actually holds for arbitrary (real or complex) zeros of arbitrary complex polynomials.