La mitjana aritmètica d'un conjunt
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{\displaystyle {x_{1},x_{,}\cdots ,n}\in \mathbb {R} ^{+}}
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{\displaystyle {\frac {x_{1}+x_{2}\cdots +x_{n}}{n}}}
La mitjana geomètrica d'un conjunt
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{\displaystyle {x_{1},x_{,}\cdots ,n}\in \mathbb {R} ^{+}}
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{\displaystyle {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}}
Siga
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{\displaystyle {x_{1},x_{,}\cdots ,n}\in \mathbb {R} ^{+}}
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{\displaystyle {\frac {x_{1}+x_{2}\cdots +x_{n}}{n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}}
Es compleix la igualtat si i només si
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{\displaystyle {x_{1}=x_{2}=\cdots =x_{n}}}
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{\displaystyle {\frac {x_{1}+x_{2}\cdots +x_{n}}{n}}={\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}}
O sigui, només són iguals la mitjana aritmètica i la mitjana geomètrica d'un conjunt de nombres positius si tots els nombres són iguals.
Aquest article o secció necessita l'atenció d'un expert en la matèria. Si us plau, ajudeu a trobar-ne un o milloreu aquesta pàgina vosaltres mateixos si podeu. (Vegeu la discussió ).
Per a demostrar la desigualtat MA-MG , es desenvolupara pel mètode d'inducció matemàtica , demostrant que la MA-MG és certa per a 2 elements, després generalitzant-lo per a 2n elements i demostrant que si certa per a n és certa per a n+1 elements.
Siga
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{\displaystyle {x_{1},x_{,}\cdots ,n}\in \mathbb {R} ^{+}}
Procedim a considerar el primer cas en què n=2
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{\displaystyle {\frac {x_{1}+x_{2}}{2}}\geq {\sqrt[{2}]{x_{1}x_{2}}}}
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{\displaystyle {\frac {(x_{1}+x_{2})^{2}}{4}}\geq x_{1}x_{2}}
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{\displaystyle (x_{1}+x_{2})^{2}\geq 4x_{1}x_{2}}
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{\displaystyle x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2}\geq 4x_{1}x_{2}}
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{\displaystyle x_{1}^{2}-2x_{1}x_{2}+x_{2}^{2}\geq 0}
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{\displaystyle (x_{1}-x_{2})^{2}\geq 0}
Quedant així demostrat per a n=2, després es demostra que si és certa per a n=2 és certa per a 2n elements.
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{\displaystyle {\frac {x_{1}+x_{2}\cdots +x_{2n}}{2n}}\geq {\sqrt[{2n}]{x_{1}x_{2}\cdots x_{2n}}}}
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{\displaystyle {\frac {{\frac {(x_{1}+x_{2}+\cdots +x_{n+1})}{n}}+{\frac {(x_{n+1}+x_{n+2}+\cdots +x_{2n})}{n}}}{2}}\geq {\sqrt[{2}]{{\frac {(x_{1}+x_{2}+\cdots +x_{n+1})}{n}}{\frac {(x_{n+1}+x_{n+2}+\cdots +x_{2n})}{n}}}}}
Seguint la hipòtesi,
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{\displaystyle {\frac {x_{1}+x_{2}\cdots +x_{n}}{n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}}
Se seguix que,
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{\displaystyle {\frac {{\frac {(x_{1}+x_{2}+\cdots +x_{n+1})}{n}}+{\frac {(x_{n+1}+x_{n+2}+\cdots +x_{2n})}{n}}}{2}}\geq {\sqrt[{2}]{{\sqrt[{n}]{(x_{1}x_{2}\cdots x_{n+1})}}{\sqrt[{n}]{(x_{n+1}x_{n+2}\cdots x_{2n})}}}}}
Sent açò igual a,
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{\displaystyle {\frac {x_{1}+x_{2}\cdots +x_{2n}}{2n}}\geq {\sqrt[{2n}]{x_{1}x_{2}\cdots x_{2n}}}}
Quedant així demostrat que si és cert per a 2 elements és cert per a 2n elements.
Ara procedim a demostrar que si és certa per a n elements és certa per a n-1 elements,
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{\displaystyle {x_{1},x_{,}\cdots ,{n-1}}\in \mathbb {R} ^{+}}
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{\displaystyle {\frac {x_{1}+x_{2}\cdots +x_{n-1}}{n-1}}}
Es considera la desigualtat de tots els elements esmentats,
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{\displaystyle {\frac {x_{1}+x_{2}+\cdots x_{n-1}+{\frac {x_{1}+x_{2}\cdots +x_{n-1}}{n-1}}}{n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n-1}{\frac {x_{1}+x_{2}\cdots +x_{n-1}}{n-1}}}}}
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{\displaystyle {\frac {(n-1)x_{1}+(n-1)x_{2}+\cdots +(n-1)x_{n-1}+x_{1}+x_{2}\cdots +x_{n-1}}{(n-1)n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n-1}}}{\sqrt[{n}]{\frac {x_{1}+x_{2}\cdots +x_{n-1}}{n-1}}}}
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{\displaystyle {\frac {nx_{1}+nx_{2}+\cdots +nx_{n-1}}{(n-1)n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n-1}}}{\sqrt[{n}]{\frac {x_{1}+x_{2}\cdots +x_{n-1}}{n-1}}}}
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{\displaystyle {\frac {x_{1}+x_{2}+\cdots +x_{n-1}}{n-1}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n-1}}}{\sqrt[{n}]{\frac {x_{1}+x_{2}\cdots +x_{n-1}}{n-1}}}}
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{\displaystyle ({\frac {x_{1}+x_{2}+\cdots +x_{n-1}}{n-1}})^{\frac {n-1}{n}}\geq ({x_{1}x_{2}\cdots x_{n-1}})^{\frac {1}{n}}}
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{\displaystyle ({\frac {x_{1}+x_{2}+\cdots +x_{n-1}}{n-1}})^{\frac {n-1}{1}}\geq ({x_{1}x_{2}\cdots x_{n-1}})}
Fent arrel (n-1)-èsima se seguix,
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{\displaystyle ({\frac {x_{1}+x_{2}+\cdots +x_{n-1}}{n-1}})\geq ({x_{1}x_{2}\cdots x_{n-1}})^{\frac {1}{^{n}-1}}}
Quedant així demostrat pel mètode inductiu, la veracitat de la desigualtat MA-MG .
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{\displaystyle {\frac {x_{1}+x_{2}\cdots +x_{n}}{n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}},\forall n\in \mathbb {N} }
Q.E.D.
![{\displaystyle {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c293138a5d3ff3a9585ca9014322ec900db1e002)
![{\displaystyle {\frac {x_{1}+x_{2}\cdots +x_{n}}{n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d8c9bfd989dc1cccbe84b8a31e442ebf96d7e32)
![{\displaystyle {\frac {x_{1}+x_{2}\cdots +x_{n}}{n}}={\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/85b4c9c9e89b90f64e817c5b3678302f4eb60836)
![{\displaystyle {\frac {x_{1}+x_{2}}{2}}\geq {\sqrt[{2}]{x_{1}x_{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4bf3bc5d87757234d17c817427c85d6c37f2a886)
![{\displaystyle {\frac {x_{1}+x_{2}\cdots +x_{2n}}{2n}}\geq {\sqrt[{2n}]{x_{1}x_{2}\cdots x_{2n}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0450ab0e5aff8a8609af26a40e37d239cd52b1f9)
![{\displaystyle {\frac {{\frac {(x_{1}+x_{2}+\cdots +x_{n+1})}{n}}+{\frac {(x_{n+1}+x_{n+2}+\cdots +x_{2n})}{n}}}{2}}\geq {\sqrt[{2}]{{\frac {(x_{1}+x_{2}+\cdots +x_{n+1})}{n}}{\frac {(x_{n+1}+x_{n+2}+\cdots +x_{2n})}{n}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b47f271dbce7303cb77866ba180013618ad87500)
![{\displaystyle {\frac {{\frac {(x_{1}+x_{2}+\cdots +x_{n+1})}{n}}+{\frac {(x_{n+1}+x_{n+2}+\cdots +x_{2n})}{n}}}{2}}\geq {\sqrt[{2}]{{\sqrt[{n}]{(x_{1}x_{2}\cdots x_{n+1})}}{\sqrt[{n}]{(x_{n+1}x_{n+2}\cdots x_{2n})}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0455083ae790487b954621e87b58a70c8f2a6414)
![{\displaystyle {\frac {x_{1}+x_{2}+\cdots x_{n-1}+{\frac {x_{1}+x_{2}\cdots +x_{n-1}}{n-1}}}{n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n-1}{\frac {x_{1}+x_{2}\cdots +x_{n-1}}{n-1}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bdfe19d3b322d66fd1840a9538618e3b9beb4e6)
![{\displaystyle {\frac {(n-1)x_{1}+(n-1)x_{2}+\cdots +(n-1)x_{n-1}+x_{1}+x_{2}\cdots +x_{n-1}}{(n-1)n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n-1}}}{\sqrt[{n}]{\frac {x_{1}+x_{2}\cdots +x_{n-1}}{n-1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ebd95c97059290a4694dc21f6eeaa61e84f01bc3)
![{\displaystyle {\frac {nx_{1}+nx_{2}+\cdots +nx_{n-1}}{(n-1)n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n-1}}}{\sqrt[{n}]{\frac {x_{1}+x_{2}\cdots +x_{n-1}}{n-1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b5e737c3d05dde9cdf498fa01c4ad5a3a2d7b17)
![{\displaystyle {\frac {x_{1}+x_{2}+\cdots +x_{n-1}}{n-1}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n-1}}}{\sqrt[{n}]{\frac {x_{1}+x_{2}\cdots +x_{n-1}}{n-1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8582d3914ca73252af7ab67623c63f4b0759348)
![{\displaystyle {\frac {x_{1}+x_{2}\cdots +x_{n}}{n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}},\forall n\in \mathbb {N} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/292749e71be7554e2e49708e9bd0553f846280f3)