$n=2:$ Let $a,b$ be positive real numbers. Prove that\[a^2+(\frac{a+b}{2})^2\le \frac{1+\sqrt{2+\sqrt{5}}}{4}(a^2+b^2).\]

sqing wrote: $n=2:$ Let $a,b$ be positive real numbers. Prove that\[a^2+(\frac{a+b}{2})^2\le \frac{1+\sqrt{2+\sqrt{5}}}{4}(a^2+b^2).\] The constant should be $\frac{3+\sqrt{5}}4$

Thank gnik. Let $a,b$ be positive real numbers. Prove that\[a^2+(\frac{a+b}{2})^2\le \frac{3+\sqrt{5}}4(a^2+b^2).\]

Let $a,b,c$ be positive real numbers. Prove that\[a^2+(\frac{a+b}{2})^2+(\frac{a+b+c}{3})^2 \le \lambda (a^2+b^2+c^2),\] where $\lambda$ be the largest real root of the equation $36{\lambda}^3-66{\lambda}^2+19{\lambda}-1=0$.

This is Hardy Inequality, and The Answer is 4

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=20811&p=135847&hilit=hardy+inequality&#p135847

YES. The answer is 4 and the equality ( well just converging to 4 actually ) case is $ x_{n}=\sqrt{n} $ It took me some time to find this equality case.

A solution hasn't been posted yet, so I will post a solution that utilizes Cauchy Balancing. Note that $ \left(\frac{1}{k}\sum_{j = 1}^{k}x_j\right)^2 \le \left(\sum_{j = 1}^{k}\frac{c_j}{k^2}\right)\left(\sum_{j = 1}^{k}\frac{x_j^2}{c_j}\right) $ for any $ c_1, c_2, \dots, c_n \in \mathbb{R}^{+}. $ Therefore \[ \sum_{k=1}^{n}\left ( \frac{1}{k}\sum_{j=1}^{k}x_j \right )^2 \le \sum_{k=1}^{n}\left(\sum_{j = 1}^{k}\frac{c_j}{k^2}\right)\left(\sum_{j = 1}^{k}\frac{x_j^2}{c_j}\right) = \sum_{k = 1}^{n}a_kx_k^2 \] Where $ a_k = \frac{c_1 + c_2 + \dots + c_k}{c_kk^2} + \frac{c_1 + c_2 + \dots + c_{k + 1}}{c_k(k + 1)^2} + \dots + \frac{c_1 + c_2 + \dots + c_n}{c_kn^2} $ for all $ k \in \{1, 2, \dots, n\}. $ By letting $ c_k = \sqrt{k} - \sqrt{k - 1} $ we find that $ a_k = \frac{1}{\sqrt{k} - \sqrt{k - 1}}\sum_{j = k}^{n}j^{-\frac{3}{2}} \le (\sqrt{k} + \sqrt{k - 1})\int_{k - 1}^{n}j^{-\frac{3}{2}} = 2(\sqrt{k} + \sqrt{k - 1})\left(\frac{1}{\sqrt{k - 1}} - \frac{1}{\sqrt{n}}\right) $ for all $ k \in \{1, 2, \dots, n\}. $ Letting $ n $ go to infinity we see that $ \boxed{c = 4} $ works and is optimal.