By playing around with assoc. Legendre polynomials, I arrived at


$$((l+1)+m) (P\_l^m(x))^2+((l+1)-m)(P\_{l+1}^m(x))^2 = 2(l+1)x P\_l^m(x)P\_{l+1}^m(x).$$


Now, I want to show that we don't have equality for $x \in (-1,1).$


I undertook quite some computations in order to be sure that this is really the case, but I currently don't see why this is true. 


There are some simple ways to start with:


For $x=0$ the inequality is obvious and both sides are even functions. 


The remarkable remark:
A remarkable, but maybe not obvious fact is that this equation is equivalent to the fact that the Wronskian of the functions $P\_l^m$ and $P\_{l+1}^m$ does not vanish inside $(-1,1)$, so if it is possible to show this, then you showed the inequality too. Maybe this has be shown somewhere, I don't know, I could not find it. 


So equivalently, I want to understand why $W(P\_l^m,P\_{l+1}^m) \neq 0$ in $(-1,1)$.



Equality cannot hold unless $x \in \{-1,1\}$ and/or Wronskian is not zero
Let $y\_1< z\_1< y\_2< z\_2< \dots< y\_{l-m}\leq z\_{l-m}< y\_{l-m+1}$ be roots of polynomials $f=(1-x^2)^{-m/2} P\_l^m$ and $g=(1-x^2)^{-m/2} P\_{l+1}^m$ ($y$'s are roots of $g$, $z$'s are roots of $f$). They are real, belong to $(-1,1)$ and alternate as written because $f$, $g$ are orthogonal polynomial in weighted $L^2\_w(-1,1)$ with weight $w=(1-x^2)^m$. 


Wronskian vanishes (in a point different from $\pm 1$) iff $(P\_{l+1}^m/P\_l^m)'=0$, i.e. iff $(g/f)'=0$, i.e. iff $f'/f=g'/g$ i.e. iff
$$
\sum \frac1{x-y\_i}=\sum \frac1{x-z\_i}.
$$
But it is impossible for real $x$. If, say, $y\_k<x<z\_k$ for some $k$, then $1/(x-y\_i)>1/(x-z\_{i-1})$ for $i=2,\dots,{l-m+1}$ and $1/(x-y\_1)>0$, thus LHS exceed RHS.


