First I will prove this for $ n \ge 4$ and leave n = 3 as a special case. for $ n \ge 4$ we will start with only the squares. $ \frac{1}{p_{1}^{2}}+\frac{1}{p_{2}^{2}}+\frac{1}{p_{3}^{2}}+\cdots+\frac{1}{p_{n}^{2}}$ Now we use the fact that $ \sum_{k=1}^{\infty}\frac{1}{p_{k}^{2}}\le \sum_{k=1}^\infty\frac{1}{k^{2}}= \frac{\pi^{2}}{6}$ and since the partial sums produce a monotone increasing sequence, we know all of them are less then the infinite sum $ \sum_{k=1}^{n}\frac{1}{p_{k}^{2}}\le \sum_{k=1}^\infty\frac{1}{p_{k}^{2}}$ Now use the sum of inverse squares to help find a low upper bound, we subtract from it 1, and all the even numbers greater then 2. $ \sum_{k=1}^\infty\frac{1}{p_{k}^{2}}\le \left(\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+\frac{1}{9^{2}}+\cdots \right) \le \frac{\pi^{2}}{6}-1-\left( \frac{1}{4^{2}}+\frac{1}{6^{2}}+\cdots \right)$ $ \frac{\pi^{2}}{6}-1-\left( \frac{1}{4^{2}}+\frac{1}{6^{2}}+\cdots \right) = \frac{\pi^{2}}{6}-1-\frac{1}{4}\left(\frac{\pi^{2}}{6}-1\right)$ $ = \frac{3}{4}\left(\frac{\pi^{2}}{6}-1\right) \thickapprox 0.4837\cdots < \frac{1}{2}$ Hence, all the partial sums for $ \frac{1}{p_{1}^{2}}+\frac{1}{p_{2}^{2}}+\frac{1}{p_{3}^{2}}+\cdots+\frac{1}{p_{n}^{2}}$ are less then one half. Now to add the extra term at the end. $ \frac{1}{p_{1}p_{2}p_{3}p_{4}}\thickapprox 0.00476\cdots$ and $ \frac{1}{p_{1}p_{2}p_{3}p_{4}}> \frac{1}{p_{1}p_{2}p_{3}p_{4}p_{5}}> \cdots > \frac{1}{p_{1}p_{2}\cdots p_{n}}$ Adding the largest terms, will give us the upper bound for all the partial sums of inverse squares plus the extra product term. $ \frac{1}{p_{1}^{2}}+\frac{1}{p_{2}^{2}}+\cdots+\frac{1}{p_{n}^{2}}+\frac{1}{p_{1}p_{2}\cdots p_{n}}< \sum_{k=1}^{\infty}\frac{1}{p_{k}^{2}}+\frac{1}{p_{1}p_{2}p_{3}p_{4}}\le \frac{\pi^{2}}{8}-\frac{3}{4}\thickapprox 0.4885 <\frac{1}{2}$ Note that this solves the problem for $ n \ge 4$ To solve for n =3, just do a simple calculation $ \frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{2.3.5}\thickapprox 0.4344\cdots <\frac{1}{2}$ Hence we have our result

i've got an easier solution but it works for $n>6$. just prove by induction that : $\sum_{i=1}^{n}\frac{1}{p_{i}^{2}}+\frac{1}{\prod_{i=1}^{n}p_{i} }<\frac{1}{2}-\frac{1}{p_{n}}$ for $n<7$ you can check it easily!!! by a lil' bit calculation!!! (sorry for this part.)

MBGO wrote: i've got an easier solution but it works for $n>6$. just prove by induction that : $\sum_{i=1}^{n}\frac{1}{p_{i}^{2}}+\frac{1}{\prod_{i=1}^{n}p_{i} }<\frac{1}{2}-\frac{1}{p_{n}}$ for $n<7$ you can check it easily!!! by a lil' bit calculation!!! (sorry for this part.) Nice solution MBGO. Would you mind to explain me how did you find that inequality because i am not so familiar with this things.Thanks