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Remember 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.

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.

Remember 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 $\{a_i\}$ is $\ge$ the $y$-norm of $\{a_i\}$. Also $x\le y$ so simple facts about $p$-norms show that LHS$\ge$HRS.

Remember 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.

Remember 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.

Remember 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 $\{a_i\}$ is $\ge$ the $y$-norm of $\{a_i\}$. Also $x\le y$ so simple facts about $p$-norms show that LHS$\ge$HRS.