freemind wrote: Let $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ be $n$ real numbers in $\left(\frac{1}{4},\frac{2}{3}\right)$. Find the minimal value of the expression: \[ \log_{\frac 32x_{1}}\left(\frac{1}{2}-\frac{1}{36x_{2}^{2}}\right)+\log_{\frac 32x_{2}}\left(\frac{1}{2}-\frac{1}{36x_{3}^{2}}\right)+\cdots+ \log_{\frac 32x_{n}}\left(\frac{1}{2}-\frac{1}{36x_{1}^{2}}\right). \] by the rearrange Inequality we have: $RHS=\log_{\frac 32x_{1}}\left(\frac{1}{2}-\frac{1}{36x_{2}^{2}}\right)+\log_{\frac 32x_{2}}\left(\frac{1}{2}-\frac{1}{36x_{3}^{2}}\right)+\cdots+ \log_{\frac 32x_{n}}\left(\frac{1}{2}-\frac{1}{36x_{1}^{2}}\right)\geq \sum_{i=1}^n {\log_{\frac 32x_{i}}{(\frac12-\frac{1}{36x_{i}^{2})}}}$ ang then we should miniumize the expression $\log_{\frac 32x}{(\frac12-\frac{1}{36x^{2}})}$ but I can't solve it. Who have the interest?

Let ${f(x)=\log_{\frac{3}{2}x}{(\frac{1}2-\frac{1}{36x^{2}})}=\log( {72x^2}\over {36x^2-2}})/\log( {2\over {3x}}).$ Taking the first derivative, and apart from some positive factors, we get the term $\log[ 2^y\cdot{{y+1}\over 8}\cdot ( 1+{1\over y})^y],$ where $y=18x^2-1.$ Clearly every term in the logarithm is an increasing function in $y,$ and the logarithm becomes zero at $y=1,$ which is in the allowed range of values for $y,$ namely $1/8<y<7.$ Thus, $y=1\leftrightarrow x=1/3$ is the only value for which $f'(x)=0.$ Checking the second derivative at $x=1/3$ we see that this value corresponds to a minimum, which is $f_{\rm min}=2.$

First we see that $\frac{3}{2}x_{i}<\frac{3}{2}\cdot\frac{2}{3}=1$. Now from AM-GM we have: $\frac{1}{4}\leqslant \frac{\frac{1}{(6x_{i})^{2}}+\frac{(6x_{i})^{2}}{16}}{2}$ so as the base of the logarithms is smaller than 1 we have: $\sum_{i=1}^{n}\log_{\frac{3}{2}x_{i}}\left(\frac{1}{2}-\frac{1}{36x_{i+1}^{2}}\right)\geqslant \sum_{i=1}^{n} \log_{\frac{3}{2}x_{i}}\left(\frac{(6x_{i+1})^2}{16}\right)$. Finally, using AM-GM we have: $2\sum_{i=1}^{n} \log_{\frac{3}{2}x_{i}}\left(\frac{6x_{i+1}}{4}\right)\geqslant 2n\sqrt[n]{\prod_{i=1}^{n}\log_{\frac{3}{2}x_{i}}\left(\frac{3}{2}x_{i+1}\right)}$. Now if we denote $\frac{3}{2}x_{i}=y_{i}$ we have: $\sum_{i=1}^{n} \log_{\frac{3}{2}x_{i}}\left(\frac{1}{2}-\frac{1}{36x_{i+1}^{2}}\right)\geqslant 2n\sqrt[n]{\frac{\prod \lg y_{i+1}}{\prod \lg y_{i}}}=2n$