Ghd wrote: If $x_1,x_2,...,x_n>0 $ and $x_1^2+x_2^2+...+x_n^2=\dfrac{1}{n}$,prove that $\sum x_i+\sum \dfrac{1}{x_i \cdot x_{i+1}} \ge n^3+1.$ n=3: If $a, b, c>0$ and $a^2+b^2+c^2=\frac{1}{3}$, prove that$$(a+b+c)(1+\frac{1}{abc}) \geq 28.$$

sqing wrote: dimacihui wrote: If $a>0, b>0, c>0$ such that $a^2+b^2+c^2=\frac{1}{3}$ prove the inequality $$(a+b+c)(1+\frac{1}{abc}) \geq 28.$$ By AM-GM , $(a+b+c)(1+\frac{1}{abc}) \geq 3\left(\frac{\sqrt[3]{abc}}{2}+\frac{\sqrt[3]{abc}}{2}+\frac{1}{54\sqrt[3]{a^2b^2c^2}}+\frac{53}{54\sqrt[3]{a^2b^2c^2}}\right)$ $\geq 3\left(\frac{1}{2}+\frac{53}{54\sqrt[3]{a^2b^2c^2}}\right)\geq 3\left(\frac{1}{2}+\frac{53}{6}\right)=28.$ Ghd wrote: If $x_1,x_2,...,x_n>0 $ and $x_1^2+x_2^2+...+x_n^2=\dfrac{1}{n}$,prove that $\sum x_i+\sum \dfrac{1}{x_i \cdot x_{i+1}} \ge n^3+1.$

We'll use the convention $x_{n+1}=x_1$ if there are only $n$ variables $x_1, x_2, \dots, x_n$. Notice that $\displaystyle \frac{1}{n}=\sum_{k=1}^{n} x_k^2 \geq n\sqrt[n]{\prod_{k=1}^{n} x_k^2} \implies \sqrt[n]{\prod_{k=1}^{n} x_k^2}\leq \frac{1}{n^2}$. So we have $$\sum_{k=1}^{n}x_k +\sum_{k=1}^{n}\frac{1}{x_kx_{k+1}} \geq n\left(\sqrt[n]{\prod_{k=1}^{n} x_k}+\frac{1}{\sqrt[n]{\prod_{k=1}^{n} x_k^2}}\right)$$$$=n\left(\frac{\sqrt[n]{\prod_{k=1}^{n} x_k}}{2}+\frac{\sqrt[n]{\prod_{k=1}^{n} x_k}}{2}+\frac{1}{2n^3\sqrt[n]{\prod_{k=1}^{n} x_k^2}}+\frac{2n^3-1}{2n^3\sqrt[n]{\prod_{k=1}^{n} x_k^2}}\right)$$$$\geq n\left(\frac{3}{2n}+\frac{2n^3-1}{2n^3\sqrt[n]{\prod_{k=1}^{n} x_k^2}}\right) \geq n^3+1.$$ Equality holds if and only if $x_1=x_2=\dots=x_n=\frac{1}{n}$.

Ghd wrote: If $x_1,x_2,...,x_n>0 $ and $x_1^2+x_2^2+...+x_n^2=\dfrac{1}{n}$,prove that $\sum x_i+\sum \dfrac{1}{x_i \cdot x_{i+1}} \ge n^3+1.$ $2LHS=2\sum_{i=1}^{n}x_{i}+2\sum_{i=1}^{n}\frac{1}{x_{i}x_{i+1}}=2\sum_{i=1}^{n}x_{i}+\sum_{i=1}^{n}\frac{1}{n^3x_{i}x_{i+1}} +\sum_{i=1}^{n}\frac{2n^3-1}{n^3x_{i}x_{i+1}}=\sum_{i=1}^{n}(x_i+x_{i+1}+\frac{1}{n^3x_ix_{i+1}})+\sum_{i=1}^{n}\frac{2n^3-1}{n^3x_{i}x_{i+1}}$ $\geqslant n(\frac{3}{n})+(2n^3-1)\frac{n^2}{n^3\sum_{i=1}^{n}xix_{i+1}}=3+(2n^3-1)\frac{1}{n\sum_{i=1}^{n}xix_{i+1}}\geqslant 3+(2n^3-1)\frac{1}{n\frac{1}{n}}=2(n^3+1)=2RHS$