{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe use the strict concavity of the logarithm.\n\n1. Let $p, q \u003e 1$ with $\\frac{1}{p} + \\frac{1}{q} = 1$, and let $a, b \\ge 0$.\n   For $c \u003e 0$ and real $r$, the [[power|0TV07AQT62]] is $c^r = e^{r \\ln c}$. For $c = 0$ and $r \u003e 0$ we use the usual convention $0^r = 0$.\n2. For $c \u003e 0$ we have $c^r \u003e 0$, because the [[exponential function is positive|0P9BNBW2W3]].\n   Also $\\ln(c^r) = r \\ln c$, because [[$\\ln$ is the inverse of $\\exp$|0RKW2X63YR]].\n3. First let $a = 0$ or $b = 0$. Then $a b = 0$, and both terms $\\frac{a^p}{p}$ and $\\frac{b^q}{q}$ are at least $0$. So the inequality holds.\n4. In this case equality means $\\frac{a^p}{p} + \\frac{b^q}{q} = 0$. Both terms are nonnegative, so this means $a^p = 0$ and $b^q = 0$.\n   One of $a^p$ and $b^q$ is already $0$. So equality holds exactly when the other one is $0$ too, that is, when $a^p = b^q$.\n5. Now let $a, b \u003e 0$. Put $x = a^p$, $y = b^q$ and $t = \\frac{1}{p}$. Then $x, y \u003e 0$ by step 2, $t \\in (0, 1)$ since $p \u003e 1$, and $1 - t = \\frac{1}{q}$.\n6. By the [[logarithm of a product|0K2FNY6DYV]] and step 2,\n   $$\\ln(a b) = \\ln a + \\ln b = \\tfrac{1}{p} \\ln x + \\tfrac{1}{q} \\ln y = t \\ln x + (1 - t) \\ln y.$$\n7. Suppose $x \\ne y$. The [[logarithm is strictly concave|0ZS54B9GJF]], so $\\ln(t x + (1 - t) y) \u003e t \\ln x + (1 - t) \\ln y = \\ln(a b)$.\n8. The [[logarithm|0044CPHJBV]] is strictly increasing by [[the inverse result|0RKW2X63YR]]. If $t x + (1 - t) y \\le a b$, then $\\ln(t x + (1 - t) y) \\le \\ln(a b)$, against step 7. So $a b \u003c \\frac{a^p}{p} + \\frac{b^q}{q}$, with strict inequality.\n9. Suppose $x = y$. Then $t x + (1 - t) y = x = a^p$. Also $\\ln b = \\frac{1}{q} \\ln y = \\frac{1}{q} \\ln x = \\frac{p}{q} \\ln a$.\n10. Hence $\\ln(a b) = \\big(1 + \\frac{p}{q}\\big) \\ln a$. Since $1 + \\frac{p}{q} = p \\big( \\frac{1}{p} + \\frac{1}{q} \\big) = p$, this is $p \\ln a = \\ln(a^p)$.\n11. The logarithm is strictly increasing, so it is injective. So $a b = a^p = \\frac{a^p}{p} + \\frac{b^q}{q}$, and equality holds.\n12. Steps 8 and 11 show that for $a, b \u003e 0$ the inequality holds, with equality exactly when $a^p = b^q$. With steps 3 and 4, this proves the statement for all $a, b \\ge 0$.\n\n## Idea\n\nTake logarithms. The logarithm of $a b$ is the weighted average of $\\ln a^p$ and $\\ln b^q$ with weights $1/p$ and $1/q$. Concavity of the logarithm says that this average is at most the logarithm of the weighted average of $a^p$ and $b^q$. Strict concavity gives the equality case.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "26D15"
    ],
    "name": "Proof of Young's inequality for products",
    "proves": "07VHQ31EYJ",
    "sources": [
      {
        "locator": "Standard version for conjugate Hölder exponents",
        "url": "https://en.wikipedia.org/wiki/Young%27s_inequality_for_products"
      }
    ],
    "tag": "4N8SCS4TTJ",
    "uses": [
      "0TV07AQT62",
      "0P9BNBW2W3",
      "0RKW2X63YR",
      "0K2FNY6DYV",
      "0ZS54B9GJF",
      "0044CPHJBV"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/4N8SCS4TTJ/",
  "links": [
    {
      "field": "proves",
      "target": "07VHQ31EYJ"
    },
    {
      "field": "uses",
      "target": "0TV07AQT62"
    },
    {
      "field": "uses",
      "target": "0P9BNBW2W3"
    },
    {
      "field": "uses",
      "target": "0RKW2X63YR"
    },
    {
      "field": "uses",
      "target": "0K2FNY6DYV"
    },
    {
      "field": "uses",
      "target": "0ZS54B9GJF"
    },
    {
      "field": "uses",
      "target": "0044CPHJBV"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/4N8SCS4TTJ.md",
  "msc": [
    "26D15"
  ],
  "name": "Proof of Young's inequality for products",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/4N8/4N8SCS4TTJ.md",
  "status": null,
  "tag": "4N8SCS4TTJ",
  "uri": "m2://4N8SCS4TTJ",
  "url": "https://math2.tamnd.com/wiki/proof-of-youngs-inequality-for-products/",
  "writing": {
    "tag": "4N8SCS4TTJ",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}