Sum of indicator function for square free integers

Question

Define $S(n)$ as $1$ when $n$ is square-free, and $0$ otherwise. Now prove that: $$\displaystyle \sum_{n \leq x}{S(n)} \geq \lfloor x\rfloor - \displaystyle\sum_{2 \leq d \leq \sqrt{x} } { \frac{x}{d^2} } $$ I have tried to prove it by induction, i.e. assume true for $x = x_0$, then consider separate cases for $x=x_0 +1 $ as follows:

1. $x_0 + 1$ is not divisible by any squares: Trivial to show inequality holds
2. $x_0 + 1$ is a square number: Trivial to show inequality holds
3. $x_0 + 1$ is not a square number but is divisible by some $d^2$.

It is the third case I am struggling with. I have also tried a counting argument but haven't come up with anything substantive. Any help would be appreciated.

Answer 1

$\left\lfloor\frac x{d^2}\right\rfloor$ is the number of multiples of $d^2$ up to $x$. If $d^2\mid a\le x$, then $d\le\sqrt x$. Thus, on the right-hand side we have the number $\lfloor x\rfloor$ of integers up $x$ and we subtract the numbers of the multiples of $d^2$ up to $x$ for all values of $d$ for which $d^2\mid a\le x$ could hold. That leaves at most the number of square-free integers up to $x$.