Đề bài: Cho: $\begin{cases}a_{1}a_{2}…a_{n}>0\left ( n\in Z,n\geq 2 \right ) \\a_{1}a_{2}+a_{2}a_{3}+…+a_{n-1}a_{n}+a_{n}a_{1}=1\\S=\sum\limits_{i=1}^n a_{i}  \end{cases}$Chứng minh rằng :$\sum\limits_{i=1}^n \frac{a_{i}^{3}}{S-a_{i}}\geq \frac{1}{n-1}$

Lời giải

Đề bài:
Cho: $\begin{cases}a_{1}a_{2}…a_{n}>0\left ( n\in Z,n\geq 2 \right ) \\a_{1}a_{2}+a_{2}a_{3}+…+a_{n-1}a_{n}+a_{n}a_{1}=1\\S=\sum\limits_{i=1}^n a_{i}  \end{cases}$Chứng minh rằng :$\sum\limits_{i=1}^n \frac{a_{i}^{3}}{S-a_{i}}\geq \frac{1}{n-1}$
Lời giải

Theo BĐT Bunhiacopski:
$1=a_{1}a_{2}+a_{2}a_{3}+…+a_{n-1}a_{n}+a_{n}a_{1}$
$\leq \sqrt{{a_{1}}^{2}+{a_{2}}^{2}+…+{a_{n}}^{2}}\sqrt{{a_{2}}^{2}+{a_{3}}^{2}+…+{a_{n}}^{2}+{a_{1}}^{2}}$
$\Rightarrow 1\leq {a_{1}}^{2}+{a_{2}}^{2}+…+{a_{n}}^{2}$  $\left ( 1 \right )$
$\left (  {a_{1}}^{2}+{a_{2}}^{2}+…+{a_{n}}^{2} \right )^{2}=$
$=\left ( \frac{{a_{1}}^{\frac{3}{2}}}{\sqrt{S-a_{1}}}.\sqrt{a_{1}\left ( S-a_{1}
\right )}+\frac{{a_{2}}^{\frac{3}{2}}}{\sqrt{S-a_{2}}}.\sqrt{a_{2}\left
( S-a_{2} \right
)}+…+\frac{{a_{n}}^{\frac{3}{2}}}{\sqrt{S-a_{n}}}.\sqrt{a_{n}\left (
S-a_{n} \right )} \right )^{2}$
$\leq \left ( \sum\limits_{i=1}^n \frac{a_{i}^{3}}{S-a_{i}} \right ).[\sum\limits_{i=1}^n a_{i}\left ( S-a_{i}
\right ) ]= \left ( \sum\limits_{i=1}^n \frac{a_{i}^{3}}{S-a_{i}} \right )[\left ( \sum\limits_{i=1}^n a_{i} \right )^{2}-\sum\limits_{i=1}^n {a_{i}}^{2}]$
$\leq\left ( \sum\limits_{i=1}^n \frac{a_{i}^{3}}{S-a_{i}} \right )\left ( n-1 \right ) \sum\limits_{i=1}^n {a_{i}}^{2} $ (vì $
( \sum\limits_{i=1}^n a_{i}  )^{2}\leq n\sum\limits_{i=1}^n {a_{i}}^{2}]$ 
$\Rightarrow \sum\limits_{i=1}^n \frac{a_{i}^{3}}{S-a_{i}}\geq \frac{{a_{1}}^{2}+{a_{2}}^{2}+…+{a^2_{n}}}{n-1}\geq \frac{1}{n-1}$ (do $\left ( 1 \right )$ )

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Chuyên mục: Bất đẳng thức Bunhiacốpxki