# A possible inequality related to binomial theorem (or, convex/concave functions)

Let $x, \ y, \ p$ be any real numbers with $x>0$, $y>0$, and $p>1$.

The question is about (most probably) an elementary inequality:

Is it always true that $x^p+y^p\leq (x+y)^p$ ?

Note that if $p$ is any positive integer, then the above inequality is obviously correct. What about if the number $p \ (\text{with} \ p>1)$ is any non-integer real number?

I guess that (by my intuition) the answer should be positive. But how can we proceed to prove this inequality?

#### Solutions Collecting From Web of "A possible inequality related to binomial theorem (or, convex/concave functions)"

Write $x=r^2 \cos^2{\theta}$, $y = r^2 \sin^2{\theta}$, for $0 \leqslant \theta \leqslant \pi/2$ (i.e. polar coordinates on the first quadrant). Then you have
$$\frac{x^p+y^p}{(x+y)^p} = \frac{r^{2p}\cos^{2p}{\theta}+r^{2p}\sin^{2p}{\theta}}{r^{2p} (1)} = \cos^{2p}{\theta} + \sin^{2p}{\theta}.$$
We clearly want to show that this is less than $1$.

Now, $0 \leqslant \cos^2{\theta},\sin^2{\theta} \leqslant 1$, and for $0 \leqslant x \leqslant 1$, we have $x^p \leqslant x$ since $p$ is at least $1$, so $x^{p-1}\leqslant 1$. Hence
$$\cos^{2p}{\theta} + \sin^{2p}{\theta} \leqslant \cos^{2}{\theta} + \sin^{2}{\theta} = 1,$$
as required.