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.

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.
