With $n\in\mathbb{N}$, what is the relation between $(x+y)^n+(x-y)^n\leq(2x)^n$ and the values of $x$ and $y$?
Or mathematically: $(x+y)^n+(x-y)^n\leq(2x)^n\iff$ the values of $x$ and $y$ are such that what?
I believe that $(x+y)^n+(x-y)^n\leq(2x)^n\implies{x}\geq{y}\geq0$.
But it's a partial answer, and I'm not even sure about that...
Thanks
Assume that $x$ is positive and set $z=\frac{y}{x}$. Then you are asking for which values of $z\in\mathbb{R}$ we have: $$ f_n(z) = (1+z)^n+(1-z)^n \leq 2^n. $$ Well, $f(z)$ is a convex even function and equality is attained in the endpoints of the interval $I=[-1,1]$, hence $f_n(z)\leq 2^n$ holds if and only if $z\in I$, i.e. $|y|\leq x$.
If $x$ is negative and $n$ is even the situation is the same, yielding $|y|\leq |x|$.
If $x$ is negative and $n$ is odd we have a concave function ($(-1-z)^n+(-1+z)^n$), yielding the reverse inequality $|y|\geq |x|$.