{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe argue by [[induction|0ASJVEFA9J]] on $n$.\nLet $x$ be a real number with $x \\ge -1$.\nFor each natural number $n$, let $P(n)$ be the claim $(1 + x)^n \\ge 1 + n x$.\nWe use the rules of the [[ordered field $\\mathbb{R}$|0QSPNASE50]].\n\n1. Base case: $(1 + x)^0 = 1$ and $1 + 0 \\cdot x = 1$. So $P(0)$ holds.\n2. Induction step: suppose that $P(n)$ holds, that is, $(1 + x)^n - (1 + n x) \\ge 0$.\n3. Since $x \\ge -1$, adding $1$ to both sides gives $1 + x \\ge 0$.\n4. The product of two nonnegative reals is nonnegative. So $\\big((1 + x)^n - (1 + n x)\\big)(1 + x) \\ge 0$ by steps 2 and 3.\n5. Expanding step 4 gives $(1 + x)^{n+1} \\ge (1 + n x)(1 + x)$.\n6. We expand: $(1 + n x)(1 + x) = 1 + (n + 1) x + n x^2$.\n7. If $x \\ge 0$, then $x^2 = x \\cdot x \\ge 0$. If $x \u003c 0$, then $-x \u003e 0$ and $x^2 = (-x)(-x) \\ge 0$. Since $n \\ge 0$, also $n x^2 \\ge 0$.\n8. Adding $1 + (n + 1) x$ to $n x^2 \\ge 0$ gives $1 + (n + 1) x + n x^2 \\ge 1 + (n + 1) x$.\n9. By steps 5, 6 and 8 and transitivity, $(1 + x)^{n+1} \\ge 1 + (n + 1) x$. So $P(n + 1)$ holds.\n10. By induction, $(1 + x)^n \\ge 1 + n x$ for every natural number $n$.\n\n## Idea\n\nEach step multiplies by $1 + x$, which is allowed in an inequality because $1 + x \\ge 0$.\nThe product $(1 + n x)(1 + x)$ equals $1 + (n + 1) x$ plus the extra term $n x^2$.\nThis extra term is never negative, so it can be dropped.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "methods": [
      "0ASJVEFA9J"
    ],
    "msc": [
      "26D05"
    ],
    "name": "Proof of Bernoulli's inequality by induction",
    "proves": "0M1MX28VXK",
    "sources": [
      {
        "locator": "Proof for integer exponent",
        "url": "https://en.wikipedia.org/wiki/Bernoulli%27s_inequality"
      }
    ],
    "tag": "ZPJYXS4MQ9",
    "uses": [
      "0ASJVEFA9J",
      "0QSPNASE50"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/ZPJYXS4MQ9/",
  "links": [
    {
      "field": "proves",
      "target": "0M1MX28VXK"
    },
    {
      "field": "uses",
      "target": "0ASJVEFA9J"
    },
    {
      "field": "uses",
      "target": "0QSPNASE50"
    },
    {
      "field": "methods",
      "target": "0ASJVEFA9J"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/ZPJYXS4MQ9.md",
  "msc": [
    "26D05"
  ],
  "name": "Proof of Bernoulli's inequality by induction",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/ZPJ/ZPJYXS4MQ9.md",
  "status": null,
  "tag": "ZPJYXS4MQ9",
  "uri": "m2://ZPJYXS4MQ9",
  "url": "https://math2.tamnd.com/wiki/proof-of-bernoullis-inequality-by-induction/",
  "writing": {
    "tag": "ZPJYXS4MQ9",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}