Looking for an inequality for $1 \leq p < \infty$

Question

Let $a_1,...,a_n$ be positive real numbers and let $0 < p < 1$. Then $$(a_1 + \cdots + a_n)^p \leq a_1^p + \cdots + a_n^p. $$ Now take $ 1 \leq p < \infty$. Can we get a similar inequality, like $$ (a_1 + \cdots + a_n)^p \leq C_p(a_1^p + \cdots + a_n^p),$$ where $C_p > 0$ is a constant only depending on $p$? If not, what is the better approach we can get?

Answer 1

Let $f(x)=x^p$, $0<p<1$ and $a_1\geq a_2\geq...\geq a_n$.

Thus, $f$ is a concave function and $(a_1+a_2+...+a_n,0,...,0)\succ(a_1,a_2,...,a_n)$.

Thus, by Karamata we obtain: $$f(a_1+a_2+...+a_n,0,...,0)\leq f(a_1)+f(a_2)+...f(a_n)$$ or $$(a_1+a_2+...+a_n)^p\leq a_1^p+a_2^p+...+a_n^p$$ and we are done!

Also, we see that we got a best estimation: $C_p=1$.

For $p\geq1$ and $a_1=a_2=...=a_n=1$ we get $C_p=n^{p-1}.$

We'll prove that $$(a_1+a_2+...+a_n)^p\leq n^{p-1}(a_1^p+a_2^p+...+a_n^p)$$ or $$\left(\frac{a_1+a_2+...+a_n}{n}\right)^p\leq\frac{a_1^p+a_2^p+...+a_n^p}{n},$$ which is Jensen for $f(x)=x^p$.