Vtyjkyuk

In a book, I saw the following:

For any two positive numbers $a$, $b$ and $p>1$, $a^1/p+b^1/p>(a+b)^1/p$. I tried to prove this but I was unsuccessful. Also, how is this inequality called, couldn't find any information online.

The inequality is false for $p <1$. For example $a^2+b^2 < (a+b)^2$ for $a, b >0$ so the inequality does not hold for $p =frac 1 2$. Suppose $p >1$. We claim that $c^t+(1-c)^t >1$ if $c,t in (0,1)$. If this is proved we can take $c =frac a a+b$ and $t =frac 1 p$ to complete the proof. To prove the claim diferentiate w.r.t. $c$ to see that the function $c to c^t+(1-c)^t$ is strictly increasing for $c <frac 1 2$ and then it strictly decreases. Hence the inequality will follow if we prove that equality holds for $c=0$ and $c=1$ which is trivial.

$a^1/p+b^1/p>(a+b)^1/p$ is false ! Consider the case $p=1/2$.

Actually, if $a,b>0$ and $p>1$, you have an equivalence between
$$
(a+b)^1/p < a^1/p + b^1/p
$$
and
$$
a+b < sum_k=0^p binompka^(p-k)/p b^k/p, \
a + b < a + b + sum_k=1^p-1 binompka^(p-k)/p b^k/p.
$$

The last inequality obviously is true and it is an equality if $a=0$ or $b=0$ or $p=1$.

Since both sides of inequality are homogeneous in $a, b$ of degree $1/p$ we can work with the assumption that $a+b=1$. Since $a, b$ are positive it follows that $0<a,b<1$ and $p>1$ so we have $$a^1/p+b^1/p>a+b=1=1^1/p=(a+b)^1/p$$