A desigualdade das médias afirma que a média aritmética é maior ou igual à média geométrica e esta maior ou igual à média harmônica.
Mais precisamente falando, seja
um conjunto não vazio de números reais positivos então:
![{\displaystyle {\frac {1}{n}}\sum _{i=1}^{n}x_{i}\geq {\sqrt[{n}]{\prod _{i=1}^{n}x_{i}}}\geq {\frac {n}{\sum _{i=1}^{n}\left({\frac {1}{x_{i}}}\right)}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9BaqoPotKOaNzBzjlCzjw1ota5ygiNaNaNago4ygw4nDBEoNBEnjm4)
onde
, veja somatório.
e
, veja produtório.
Queremos mostrar que:
![{\displaystyle {\frac {x_{1}+x_{2}}{2}}\geq {\sqrt {x_{1}\cdot x_{2}}}\geq {\frac {2}{{\frac {1}{x_{1}}}+{\frac {1}{x_{2}}}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9Ez2iOztlEyjG5ygw3atnFatK1ytlFaDw4ntaQoNG4atnFzghAajo4)
Como
e
são reais, temos:
![{\displaystyle (x_{1}-x_{2})^{2}\geq 0}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8NoqaPoNoPatGNajo4nDiQajdDateQzqnAytBCnjo3atGNzDe1zArE)
Expandindo, temos:
![{\displaystyle x_{1}^{2}-2x_{1}x_{2}+x_{2}^{2}\geq 0}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9EoNGNyjKPnqsOnAs5ntzCoDrEyqoOzNs0oqzFoqdFygaQaAe1aDrC)
Somando
, obtemos:
![{\displaystyle x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2}\geq 4x_{1}x_{2}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9DzAo0yqe0yga5zDm4zgzDotm4ytaNzDa4nDC5otoQo2vEzDhDaDoO)
Assim:
![{\displaystyle (x_{1}+x_{2})^{2}\geq 4x_{1}x_{2}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO83oNo2nAo2oAo3oAhEateQnghEotwQoDG5oDm2aDa3aAaOztaPoNs2)
Assumindo como sendo números positivos, podemos tomar a raiz quadrada e dividir por 2:
![{\displaystyle {\frac {x_{1}+x_{2}}{2}}\geq {\sqrt {x_{1}x_{2}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO81oAhAntnEz2a2a2vAoDvDaDCQoDdDaqoQaja2oNCPyjzFaDzBotm1)
A primeira desigualdade segue. Para mostrar a segunda, escreva esta última como:
![{\displaystyle {\frac {2}{x_{1}+x_{2}}}\leq {\frac {1}{\sqrt {x_{1}x_{2}}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8QaDFDoNeOztKPnDiOoNvEyjs1aAvAoNmNaDJDzAs0oto0zDvFzDC1)
Multiplique ambos os lados por :
:
![{\displaystyle {\frac {2x_{1}x_{2}}{x_{1}+x_{2}}}\leq {\sqrt {x_{1}x_{2}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9FnqzAygvAaNsOoDe0zjKNzDmPzAe4yjs0o2zCyte3otdDnjFBytCQ)
E observe que esta é justamente a desigualdade que procuramos, pois:
![{\displaystyle {\frac {2x_{1}x_{2}}{x_{1}+x_{2}}}={\frac {2}{\frac {x_{1}+x_{2}}{x_{1}x_{2}}}}={\frac {2}{{\frac {1}{x_{1}}}+{\frac {1}{x_{2}}}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO83zNwNntw5aDhBz2hDaNzAaAwPaqvFzjCPoNrDzge4nqeNzNBFa2w1)
E o resultado segue.
Demonstração no caso
[editar | editar código-fonte]
Queremos a igualdade para
, com k inteiro positivo.
Procederemos por indução em k:
O caso k=1, já foi demonstrado.
Suponha então que a desigualdade é valida para um certo k positivo, escreva para
:
![{\displaystyle {\frac {1}{2n}}\sum _{i=1}^{2n}x_{i}={\frac {1}{2n}}\left[\sum _{i=1}^{n}x_{i}+\sum _{i=n+1}^{2n}x_{i}\right]}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO81ajs0nje5oAdAz2zCoqs4zDBEyqzCygwOntw2nDo3ntFFoDmOntGQ)
Aplique a desigualdade da média com dois elementos:
![{\displaystyle {\frac {1}{2n}}\sum _{i=1}^{2n}x_{i}\geq {\sqrt {\left(\sum _{i=1}^{n}x_{i}\right)\cdot \left(\sum _{i=n+1}^{2n}x_{i}\right)}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8PngvByge3zNoNzDi4oNvAzjC5nAwPygvBaNdFzNo0yjsNnqhAajK1)
Agora, aplique a desigualdade para n elementos em cada um dos termos:
![{\displaystyle {\frac {1}{2n}}\sum _{i=1}^{2n}x_{i}\geq {\sqrt {{\sqrt[{n}]{\prod _{i=1}^{n}x_{i}}}\cdot {\sqrt[{n}]{\prod _{i=n+1}^{2n}x_{i}}}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO84ajGQntzAzNiQa2e0oNJFyja1aAvCzAiOaDGNnDG4aNhBzjiOzDBB)
E assim, conclua:
![{\displaystyle {\frac {1}{2n}}\sum _{i=1}^{2n}x_{i}\geq {\sqrt[{2n}]{\prod _{i=1}^{2n}x_{i}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8Oa2a4aNCNzjzAygoNaNrEoDi0zAs2ago3a2eQaje0yqhBo2eNotrA)
E a primeira desigualdade segue pois
Usemos o mesmo procedimento para demonstrar a segunda desigualdade:
![{\displaystyle {\sqrt[{2n}]{\prod _{i=1}^{2n}x_{i}}}={\sqrt {{\sqrt[{n}]{\prod _{i=1}^{n}x_{i}}}\cdot {\sqrt[{n}]{\prod _{i=n+1}^{2n}x_{i}}}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9AzjoPyjoQztG3oDsNnjzCzge0ygo1nDKNatoOnjw1nAe2ote4aAvB)
![{\displaystyle {\sqrt[{2n}]{\prod _{i=1}^{2n}x_{i}}}\geq {\frac {2}{{\frac {1}{\sqrt[{n}]{\displaystyle \prod _{i=1}^{n}x_{i}}}}+{\frac {1}{\sqrt[{n}]{\displaystyle \prod _{i=n+1}^{2n}x_{i}}}}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO84oNeNzgvCythAzNGOoNoOygdByjdAats5ajG5nAnEz2sNats3ats4)
![{\displaystyle {\sqrt[{2n}]{\prod _{i=1}^{2n}x_{i}}}\geq {\frac {2n}{\displaystyle \sum _{i=1}^{n}{\frac {1}{x_{i}}}+\displaystyle \sum _{i=n+1}^{2n}{\frac {1}{x_{i}}}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9FzjeQoNGPnjiOoAa2aja5agnEatG1zqaOzti1ngi5nDBDnjs2ajwO)
![{\displaystyle {\sqrt[{2n}]{\prod _{i=1}^{2n}x_{i}}}\geq {\frac {2n}{\displaystyle \sum _{i=1}^{2n}{\frac {1}{x_{i}}}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO82ngw2yjG0ytsPotlBaDmNzgoQate3zqw0a2nEnDKPyge5oArDoDi0)
E a segunda desigualdade segue.
Completaremos a demonstração, mostrando que se a desigualdade for válida para n termos, então também é válida para n-1 termos.
Suponha, então, que a desigualdade é válida para um número inteiro n maior que 1, ou seja:
![{\displaystyle {\frac {1}{n}}\sum _{i=1}^{n}x_{i}\geq {\sqrt[{n}]{\prod _{i=1}^{n}x_{i}}}\geq {\frac {n}{\sum _{i=1}^{n}\left({\frac {1}{x_{i}}}\right)}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9BaqoPotKOaNzBzjlCzjw1ota5ygiNaNaNago4ygw4nDBEoNBEnjm4)
Escreva:
![{\displaystyle p={\frac {1}{n-1}}\sum _{i=1}^{n-1}x_{i}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9Co2a3nDs0ntG2zNzCaAs2ztC3nAwQnqs3ytm1ntK4ztaQnqwPoNeP)
![{\displaystyle q={\sqrt[{n-1}]{\prod _{i=1}^{n-1}x_{i}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO8OatePnto1a2nBztK0oAa3njhBzqo1ajmQytlAoti2agrBaNFEytnF)
![{\displaystyle r={\frac {n-1}{\displaystyle \sum _{i=1}^{n-1}\left({\frac {1}{x_{i}}}\right)}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9DyjeNoNsNztJCatzCaDK4nAdDats1zjlCzDJAntw3ztmOoAhDzDwO)
Queremos mostrar que
Substitua
![{\displaystyle {\frac {1}{n}}\left(\sum _{i=1}^{n-1}x_{i}+q\right)\geq {\sqrt[{n}]{q\prod _{i=1}^{n-1}x_{i}}}\geq {\frac {n}{\sum _{i=1}^{n-1}\left({\frac {1}{x_{i}}}\right)+{\frac {1}{q}}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO80oAw5yqnFaNhBatFByjK4ygs5zNePaNJDajdCzqhAzjJDaNKOzqrD)
Observe que:
![{\displaystyle {\sqrt[{n}]{q\prod _{i=1}^{n-1}x_{i}}}={\sqrt[{n}]{qq^{n-1}}}=q}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9Eo2s1zjdEataPzNJBnji3zAiNngzDots5aNGPaDiNztrCngw5ajvA)
Assim temos, da primeira desigualdade:
![{\displaystyle {\frac {1}{n}}\left(\sum _{i=1}^{n-1}x_{i}+q\right)\geq q}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9Faga3ngnAaji3aNiPytm2otm1ztFCaNi5zNm4yqe4ytw4o2iNyjKQ)
Rearranjando, temos:
![{\displaystyle p={\frac {1}{n-1}}\sum _{i=1}^{n-1}x_{i}\geq q}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9AzjlEo2a5nDhDzjs3zjo2ntsNzjlFzDs0nto4zga3yta5yqi1z2aQ)
A segunda desigualdade diz:
![{\displaystyle q\geq {\frac {n}{\sum _{i=1}^{n-1}\left({\frac {1}{x_{i}}}\right)+{\frac {1}{q}}}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9EzNK2zAs5oAdBngo1aNwQzjhEaDhCnDm4agi3nDzCoNlCntzBaDoN)
O que equivale a:
![{\displaystyle \sum _{i=1}^{n-1}\left({\frac {1}{x_{i}}}\right)+{\frac {1}{q}}\geq {\frac {n}{q}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO85zghEotvEygs1njKNzjC4zjsPzge2atK0oto2nto1oto2ytK1nti5)
ou:
![{\displaystyle \sum _{i=1}^{n-1}\left({\frac {1}{x_{i}}}\right)\geq {\frac {n-1}{q}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO9DzqhFnqhDzqs4zArBaqzEzgsPoqaOzAwNntw4ntzAoNFEoNvDnqzF)
Equivalente a:
![{\displaystyle q\geq {\frac {n-1}{\sum _{i=1}^{n-1}\left({\frac {1}{x_{i}}}\right)}}=r}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO85zqw4a2w2a2wOzqdFytCPoDvFajsNaAzEnqi0nDlBzgdDoDa2nAo1)
O que completa a demonstração.
Desigualdade entre as Médias Quadrática e Aritmética[editar | editar código-fonte]
Se, na desigualdade de Cauchy fizermos
, ela assume a forma:
≤![{\displaystyle \scriptstyle {\sqrt {{a_{1}}^{2}+{a_{2}}^{2}+{a_{3}}^{2}+...+{a_{n}}^{2}}}\scriptstyle {\sqrt {n}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO85aNC5oNGOoAs5aqzFajw4oDs5zqdBateQoNsNzNhEygaOaAa5zqs1)
- Agora é só dividir os membros da desigualdade acima por
.
- Finalmente:
≥![{\displaystyle {\frac {a_{1}+a_{2}+a_{3}+...+a_{n}}{n}}}](https://amansaja.com/index.php?q=Mfv0Kfa6bO93MqTXLqrCMqiSL3dZb2hQMu9Onpz0p3oPb21BngBFb21FJgGRKArSngrOb3z2nO80oqaOntBFajdBo2o3aqaNaghCajo1ajC3oArEatFEytrAyqdData3)