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  "body": "\n## Proof\n\nWe normalize both sides and apply Young's inequality to each term.\n\n1. Let $S = \\sum_{i=1}^{n} |a_i|^p$ and $T = \\sum_{i=1}^{n} |b_i|^q$. Let $A = S^{1/p}$ and $B = T^{1/q}$, with the [[real power|0TV07AQT62]] $c^r = e^{r \\ln c}$ for $c \u003e 0$ and the convention $0^r = 0$ for $r \u003e 0$.\n2. For $c \u003e 0$ we have $c^p \u003e 0$, because the [[exponential function is positive|0P9BNBW2W3]].\n3. Suppose $S = 0$. Each term $|a_i|^p$ is at least $0$, so each term is $0$. By step 2, each $a_i = 0$.\n   Then the left side is $0$, and the right side is at least $0$. So the inequality holds. The case $T = 0$ is the same.\n4. From now on let $S \u003e 0$ and $T \u003e 0$. Then $A = e^{\\frac{1}{p} \\ln S} \u003e 0$, and $\\ln A = \\frac{1}{p} \\ln S$ since [[$\\ln$ is the inverse of $\\exp$|0RKW2X63YR]].\n   So $A^p = e^{p \\ln A} = e^{\\ln S} = S$. In the same way $B \u003e 0$ and $B^q = T$.\n5. Let $c \\ge 0$. We claim $(c / A)^p = c^p / A^p$. For $c = 0$ both sides are $0$.\n   For $c \u003e 0$, the [[logarithm of a quotient|0J7PHRBNNS]] gives $(c/A)^p = e^{p \\ln c - p \\ln A}$. By the [[exponential of a sum|0E8HNVBCSP]], $e^{p \\ln c - p \\ln A} \\, e^{p \\ln A} = e^{p \\ln c}$. So $(c/A)^p = c^p / A^p$.\n6. In the same way, $(c / B)^q = c^q / B^q$ for each $c \\ge 0$.\n7. Put $x_i = |a_i| / A$ and $y_i = |b_i| / B$. These are nonnegative. By [[Young's inequality|07VHQ31EYJ]] and steps 4 to 6,\n   $$x_i y_i \\le \\frac{x_i^p}{p} + \\frac{y_i^q}{q} = \\frac{|a_i|^p}{p S} + \\frac{|b_i|^q}{q T}.$$\n8. Add these inequalities over $i = 1, \\dots, n$. The right sides add up to $\\frac{S}{p S} + \\frac{T}{q T} = \\frac{1}{p} + \\frac{1}{q} = 1$.\n9. Since [[$|a_i b_i| = |a_i| |b_i|$|0GMV7GP8W0]], the left sides add up to $\\frac{1}{A B} \\sum_{i=1}^{n} |a_i b_i|$. So this sum is at most $1$.\n10. Multiply by $A B \u003e 0$. This gives $\\sum_{i=1}^{n} |a_i b_i| \\le A B = \\big( \\sum_{i=1}^{n} |a_i|^p \\big)^{1/p} \\big( \\sum_{i=1}^{n} |b_i|^q \\big)^{1/q}$.\n\n## Idea\n\nBoth sides scale in the same way when all $a_i$ or all $b_i$ are multiplied by a positive number. So we may scale the two sums on the right to equal $1$. Then Young's inequality bounds each product $|a_i| |b_i|$, and the bounds add up to $\\frac{1}{p} + \\frac{1}{q} = 1$.\n",
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