Show that $2^{n-1}(x^n + y^n ) \geqslant (x+y)^n $ Let $x,y$ be real numbers such that $x+y\geqslant 0$. Prove that $2^{n-1}(x^n + y^n ) \geqslant (x+y)^n $ for every $n \in N$. Can we prove this using only inequalities,without using induction?
 A: Unfortunately, you cannot use the (Jensen's) inequality for the convex function $f(x)=x^n$ directly here, because you would need to consider this function for both positive and negative values of $x$ and $y$, and the function is not convex for negative values if $n$ is odd. (Consider, for example, $n=3$ and $x=-y$ to see why the convexity cannot be used here.)
Consider $u=\frac{x+y}{2}$ and $v=\frac{x-y}{2}$. Then, we need to prove that
$2^{n-1}((u+v)^n+(u-v)^n)\ge 2^n u^n$
where $u$ is non-negative. So, the result is immediate:
$(u+v)^n+(u-v)^n=2u^n+2\binom{n}{2}u^{n-2}v^2+\ldots \ge 2u^n$.
BTW, from here you can see that the equality holds iff $x=y$ ($v=0$) or $x=-y$ and $n$ is odd ($u=0$ and the last term on the left has $u$ in it).
A: Equivalently
$$
\frac{x^n+y^n}{2}\ge\left(\frac{x+y}{2}\right)^n,
$$
i.e.,
$$
f\left(\frac{x+y}{2}\right)\le\frac{f(x)+f(y)}{2},
$$
which holds whenever $f$ is convex, and $f(x)=x^n$ is indeed convex for $n\ge 1$, as
$$f''(x)=n(n-1)x^{n-2}\ge 0,$$ for all $x\ge 0$.  
