Daca \( a_1,\ a_2,\ \dots ,\ a_n > 0 \) cu \( a_1^{2}+a_2^{2}+\dots+a_n^{2}=1 \), atunci:
\( a_1+a_2+\dots+a_n+\frac{1}{a_1a_2... a_n}\ge \sqrt{n}+\sqrt{n^n} \)
Inegalitate conditionata in n variabile
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Mateescu Constantin
- Newton
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- Location: Pitesti
-
Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
\( a_1+a_2+...+a_n+\frac{1}{a_1a_2...a_n}=(a_1+a_2+...+a_n)(a_1^2+a_2^2+...+a_n^2)^{\frac{n-1}{2}}+\frac{1}{a_1a_2...a_n}\ge n\sqrt[n]{a_1a_2...a_n}(n\sqrt[n]{a_1a_2...a_n})^{\frac{n-1}{2}}+\frac{1}{a_1a_2...a_n}=n^{\frac{n-1}{2}}p+\frac{1}{p},\ \mbox{unde}\ p=a_1a_2...a_n \)
Astfel ramane de aratat ca \( n^{\frac{n-1}{2}}p+\frac{1}{p}\ge \sqrt{n}+\sqrt{n^n} \)
\( \Longleftrightarrow (1-p\sqrt{n})(1-p\sqrt{n^n})\ge 0 \), inegalitate care este adevarata deoarece avem \( 1=a_1^2+a_2^2+...+a_n^2\ge n\sqrt[n]{a_1^2a_2^2...a_n^2}=np^{\frac{2}{n}}. \). Egalitatea are loc pentu \( a_1=a_2=...=a_n=\frac{1}{\sqrt{n^n}}. \)
Astfel ramane de aratat ca \( n^{\frac{n-1}{2}}p+\frac{1}{p}\ge \sqrt{n}+\sqrt{n^n} \)
\( \Longleftrightarrow (1-p\sqrt{n})(1-p\sqrt{n^n})\ge 0 \), inegalitate care este adevarata deoarece avem \( 1=a_1^2+a_2^2+...+a_n^2\ge n\sqrt[n]{a_1^2a_2^2...a_n^2}=np^{\frac{2}{n}}. \). Egalitatea are loc pentu \( a_1=a_2=...=a_n=\frac{1}{\sqrt{n^n}}. \)