Show that $\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq\frac{3n-2}{2n-1}$. Let $$x_n=\frac{1}{1^2}+\frac{1}{2^2}+\ldots+\frac{1}{n^2},\;\forall n\geq 1.$$ Show that $$\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq \frac{3n-2}{2n-1}.$$ 
I tried to prove it by induction  but I got a contradiction.
 A: It is trivial if you use $$\sum_{n=1}^\infty\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$$
Note that $x_1=1$ and $x_i\ge\dfrac{5}{4}\forall i\ge 2$. Hence, for $n\ge 4$,
$$LHS = \sum_{i=1}^n\dfrac{1}{i^2x_i^2}$$
$$\le1+\dfrac{16}{25}\sum_{i=2}^n\dfrac{1}{i^2}$$
$$<1+\dfrac{16}{25}\left(\dfrac{\pi^2}{6}-1\right)$$
$$<1.42<RHS$$
(using the fact that RHS is an increasing sequence which takes value $\dfrac{10}{7}$ at $n=4$). Verify for $n=1,2,3$ and we are done.
A: First note that 
\begin{align}x_k&\geq 1+\frac1{2\cdot 3}+\frac{1}{3\cdot 4}+\ldots+\frac{1}{k\cdot(k+1)}\\&=1+\frac{1}{2}-\frac{1}{k+1}=\frac{3}{2}-\frac{1}{k+1}=\frac{3k+1}{2(k+1)}.\end{align}
Therefore,
$$k^2x_k^2\geq \frac{k^2(3k+1)^2}{4(k+1)^2}.$$
For $k\ge 2$,
$$\frac{1}{k^2x_k^2}\leq \frac{4(k+1)^2}{k^2(3k+1)^2}<\frac{4}{9}\left(\frac{1}{k-7/6}-\frac{1}{k-1/6}\right).$$
This is because
\begin{align}4k^2(3k+1)^2-36(k+1)^2(k-7/6)(k-1/6)&=\frac{(57k+17)^2-688}{57}\\&\geq\frac{131^2-688}{57}>0. \end{align}
Therefore for $n\ge 2$,
\begin{align}\sum_{k=1}^n\frac{1}{k^2x_k^2}&=1+\frac{1}{5}+\sum_{k=3}^n\frac{1}{k^2x_k^2}\le 1+\frac15+\frac{4}{9}\sum_{k=3}^n\left(\frac{1}{k-7/6}-\frac{1}{k-1/6}\right)
\\&\le 1+\frac{1}{5}+\frac{4}{9}\left(\frac{1}{3-\frac76}-\frac{1}{n-\frac16}\right)=\frac{238}{165}-\frac{8}{3(6n-1)}\\&=\frac{2(238n-113)}{55(6n-1)}<\frac{3n-2}{2n-1}\end{align}
because
\begin{align}55(6n-1)(3n-2)-2(238n-113)(2n-1)&=\frac{(76n+103)^2-28241}{152}
\\&\geq \frac{255^2-28241}{152}>0.\end{align}
For $n=1$, the inequality to be proven is an equality.  Therefore, the desired ineq holds for all positive integers $n$.
