<s>[INST] If $P_n$ is $n$th Legendre polynomial and $|x|\leqslant1$ then $\frac{1-x^2}{n^2}(P'_n)^2+P^{2}_{n} \le 1$ I am trying to prove theorem 17 (h) of chapter 7 from Kaplan's Advanced Calculus book. Being $P_{n}$ a legendre polynomial and $P^{\prime}_{n}$ its derivative, I need to prove: 
$$\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} \le 1 \hspace{1cm} (n \ge 1 , |x| \le 1)$$
I already proved that (see scholia):
$\tag{1}\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} = \frac{1-x^2}{n^2}P^{\prime 2}_{n-1}+P^{2}_{n-1}$
So I tried to do a "recursive prove":
$$\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} \le \frac{1-x^2}{(n-1)^2}P^{\prime 2}_{n-1}+P^{2}_{n-1} \le \frac{1-x^2}{(n-2)^2}P^{\prime 2}_{n-2}+P^{2}_{n-2} \le ... \le (1-x^2)P^{\prime 2}_{1}+P^{2}_{1}$$
And as:
$$(1-x^2)P^{\prime 2}_{1}+P^{2}_{1} = (1-x^2) +x^2 = 1$$
It seems to me that:
$$\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} \le 1 \hspace{1cm} (n \ge 1)$$
Should be valid for any value of $x$. Not only for $|x| \le 1$. I am afraid I am missing some part of the whole history here. Can anyone help me?
Scholia:
As some people doubted (1) I will show the prove for it.
As cited in Kaplan's book the Legendre polynomials have these two properties:
$\tag{2}P^{\prime}_{n}(x) = xP^{\prime}_{n-1}(x)+nP_{n-1}(x)$
And
$\tag{3}P_{n}(x) = \frac{(x^{2}-1)}{n}P^{\prime}_{n-1}(x)+xP_{n-1}(x)$
So we can substitute (2) and (3) on the LHS of (1) and get:
$$\frac{1-x^2}{n^2}P^{\prime 2}_{n}+P^{2}_{n} = \frac{(1-x^2)}{n^2}(x^{2}P^{\prime 2}_{n-1} + 2xnP^{\prime}_{n-1}P_{n-1} + n^{2}P^{2}_{n-1}) + \frac{(x^{2}-1)^2}{n^2}P^{\prime 2}_{n-1}+ 2x\frac{(x^{2}-1)}{n}P^{\prime}_{n-1}P_{n-1}+x^{2}P^{2}_{n-1}$$
$$\frac{P^{\prime 2}_{n-1}}{n^2}(-x^{4}+x^{2}+x^{4}-2x^{2}+1)+P^{2}_{n-1}(1-x^{2}+x^{2}) = \frac{1-x^2}{n^2}P^{\prime 2}_{n-1}+P^{2}_{n-1}$$
And we get (1). The properties (2) and (3) I used in order to demonstrate (1) can be demonstrated using Rodrigues formula and Leibnitz product rule for $n$th order derivatives. I can provide the step-by-step if needed.
 [/INST] Given the identity
$$\begin{eqnarray*} \frac{1-x^2}{n^2}P_n'^2+P_n^2 &=& \frac{1-x^2}{n^2}P_{n-1}'^2+P_{n-1}^2\\&=&\left(1-\frac{1}{n}\right)^2\left(\frac{1-x^2}{(n-1)^2}P_{n-1}'^2+P_{n-1}^2\right)+\left(1-\left(1-\frac{1}{n}\right)^2\right) P_{n-1}^2\\\\&\stackrel{\text{Ind.Hyp.}}{\leq}&\left(1-\frac{1}{n}\right)^2+\left(1-\left(1-\frac{1}{n}\right)^2\right) P_{n-1}^2\end{eqnarray*}$$
($\text{Ind.Hyp}$ stands for Inductive Hypothesis) the claim is straightforward, since $P_{n-1}(x)^2$ attains its maximum value over $[-1,1]$ at the endpoints of such interval, and such maximum value is $1$.
The inequality can be refined for $x$-points belonging to a neighbourhood of the origin, since as shown here we have
$$ P_n(x)^2\leq \frac{4}{\pi(2n+1)}\cdot\frac{1}{\sqrt{1-x^2}}.$$
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