You have

$$EX=\int_0^{\infty}xdF(x)$$

Notice that $dF(x)=d(1-F(x))$ and that $P(X>t)=1-F(t)$ and use integration by parts.

Now show that for monotone decreasing positive function

$$\sum_{n=0}^\infty f(n)\ge\int_0^{\infty} f(t) dt$$

Combine these two results and you get your desired result. Hint for the second, recall Riemman sums.

You have

$$EX=\int_0^{\infty}xdF(x)$$

Notice that $dF(x)=-d(1-F(x))$ and that $P(X>t)=1-F(t)$ and use integration by parts.

Now show that for monotone decreasing positive function

$$\sum_{n=0}^\infty f(n)\ge\int_0^{\infty} f(t) dt$$

Combine these two results and you get your desired result. Hint for the second, recall Riemman sums.