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.