Let $ a_{1},a_{2},...a_{n}\in (0,\infty ) $ such that $ \sum\limits_{i=1}^{n}a_{i}=1,n\geq 2. $ Prove that:
$$ \frac{log_{a_{1}}^{2}a_{2}}{na_{1}+n-1}+\frac{log_{a_{2}}^{2}a_{3}}{na_{2}+n-1}+...+\frac{log_{a_{n}}^{2}a_{1}}{na_{n}+n-1}\geq 1. $$
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1 Answer
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First we write $\log_{a_i}a_j=\frac{\ln(a_j)}{\ln(a_j)}$ for $i\neq j$ and $i,j\in \{1,...,n\}$. Then by Titu Andreescu's inequality we have that:
$\frac{(\log_{a_1}a_2)^2}{na_1+n-1}+...+\frac{(\log_{a_n}a_1)^2}{na_n+n-1}\geq \frac{(\frac{\ln(a_1)}{\ln(a_2)}+...+\frac{\ln(a_n)}{\ln(a_1)})^2}{n(a_1+...+a_n)+n(n-1)}=\frac{(\frac{\ln(a_1)}{\ln(a_2)}+...+\frac{\ln(a_n)}{\ln(a_1)})^2}{n^2}\ (1)$
Since the sum of the positive $a_1,...,a_n$ is $1$, we have that:
$0<a_1,...,a_n<1\Rightarrow \ln(a_1),...,\ln(a_n)<0\Rightarrow \frac{\ln(a_j)}{\ln(a_j)}>0$ for $i\neq j$ and $i,j\in \{1,...,n\}$.
Thus we can apply the AM-GM inequality and get:
$\frac{\ln(a_1)}{\ln(a_2)}+...+\frac{\ln(a_n)}{\ln(a_1)}\geq n\ (2)$
If we combine $(1)$ and $(2)$, we conclude the desired inequality.
