A question about CLT for infinite variance

This is an example from Durrett (Page 131, 5th edition, Durrett). Suppose $$X_1,X_2,..$$ are i.i.d. and have $$P(X_1>x) = P(X_1 and $$P(|X_1|>x)=x^{-2}$$ for $$x\geq 1.$$ $$E|X_1|^2 = \infty$$ but $$S_n = X_1+..X_n$$ suitably normalized converges to a normal dist.

In the proof, I have one question.

Let $$Y_{n,m} = X_m 1_{(|X_m|\leq n^{1/2}\log\log{n})}$$ and $$c_n = n^{1/2}\log\log{n}$$. To show $$EY_{n,m}^2 \geq \log{n}$$, it says

$$P(|Y_{n,m}|>x) = P(|X_1|>x)-P(|X_1|>c_n)$$ $$\geq (1-(\log\log{n})^{-2})P(|X_1|>x)$$ when $$x\leq \sqrt{n}$$ which I don't understand why because that means $$P(|X_1|>c_n) \leq (\log\log{n})^{-2}P(|X_1|>x)$$ which I don't know why is true?

Thanks.

Assuming that $$n$$ is large enough in order to ensure that $$c_n\geq 1$$, we have $$\Pr(\lvert X_1\rvert>c_n)=c_n^{-2}= \frac 1n \frac 1{(\log\log n)^2}$$ and since $$x\leq \sqrt n$$, then $$n\geq x^2$$ hence $$\Pr(\lvert X_1\rvert>c_n) \leq \frac 1{x^2} \frac 1{(\log\log n)^2}$$ which is exactly what we want.