2
$\begingroup$

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:

$$ \frac{log_{a_{1}}^{2}a_{2}}{na_{1}+n-1}+\frac{log_{a_{2}}^{2}a_{3}}{na_{2}+n-1}+...+\frac{log_{a_{n}}^{2}a_{1}}{na_{n}+n-1}\geq 1. $$

$\endgroup$

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.