Conjecture: Let $a_1, a_2, \cdots , a_n>0$ and $y \ge x $ then
$$(a_1^x+a_2^x+\cdots+a_n^x)^y \ge (a_1^y+a_2^y+\cdots+a_n^y)^x$$
Equality iff $x=y$
Is the conjecture right? Have you ever seen this inequality before?
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1$\begingroup$ Do you assume $x>0$? $\endgroup$– LucenapositionCommented Aug 26 at 3:42
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2$\begingroup$ Here is an elementary proof. First by change of variables reduce the problem to $(\sum a_i)^{(y/x)}\geq \sum a_i^{(y/x)}$, then reduce it to showing $(a+b)^{(p/q)}\geq a^{(p/q)}+b^{(p/q)}$ where $p$ and $q$ are integers and $p\geq q$, then reduce it to showing $(1+a^q)^p\geq (1+a^p)^q$ where $a \leq 1$. Now this is obvious since both base and exponent of one side is larger. $\endgroup$– user127776Commented Aug 26 at 4:12
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1 Answer
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Here is a proof that works when $y\ge x>0$:
Note that $$\left(\sum_{i=1}^na_i^x\right)^y\ge\left(\sum_{i=1}^na_i^y\right)^x$$ iff $$\left(\sum_{i=1}^na_i^x\right)^{1/x}\ge\left(\sum_{i=1}^na_i^y\right)^{1/y}$$ by taking $xy$-th root. So we just need to show the $x$-norm of $\{a_i\}_{1\le i\le n}$ is ≥ the $y$-norm of $\{a_i\}_{1\le i\le n}$. Also $x\le y$ so some facts about $p$-norms show that the second inequality is true.
