# Why $\left(\sum_{p\leq \sqrt x}\frac{1}{p}\right)^2\leq \sum_{p_1p_2\leq x}\frac{1}{p_1p_2}\leq\left(\sum_{p\leq x}\frac{1}{p}\right)^2$?

Could someone tell me why $$\left(\sum_{p\leq \sqrt x}\frac{1}{p}\right)^2\leq \sum_{p_1p_2\leq x}\frac{1}{p_1p_2}\leq\left(\sum_{p\leq x}\frac{1}{p}\right)^2$$ is true ? I recall that $p$, $p_1$ and $p_2$ are prime numbers.

I was thinking to Cauchy-Schwartz, but it doesn't look to work.

One can notice that$$\left(\sum_{p\leq \sqrt x}\frac{1}{p}\right)^2=\sum_{p_1\leq\sqrt x,p_2\leq \sqrt x}\frac{1}{p_1p_2}\leq\sum_{p_1p_2\leq x}\frac{1}{p_1p_2}$$since every tuple $(p_1,p_2)$ satisfying $p_1\leq\sqrt x,p_2\leq \sqrt x$ also satisfies $p_1p_2\leq x$. Also, $$\sum_{p_1p_2\leq x}\frac{1}{p_1p_2}\leq\sum_{p_1\leq x,p_2\leq x}\frac{1}{p_1p_2}=\left(\sum_{p_1\leq x}\frac{1}{p_1}\right)\left(\sum_{p_2\leq x}\frac{1}{p_2}\right)=\left(\sum_{p\leq x}\frac{1}{p}\right)^2$$where the inequality came from the fact that $p_1p_2 \leq x$ implies both $p_1\leq x$ and $p_2\leq x$.