integration - Expectation for a skewed distribution

Let $X$ be a continuous random variable with distribution $F(x)=P(X\leq x)$ and support $[0,\infty)$. Suppose for a fixed $X_0$ and any $t\geq 0, P(X\geq X_0 +t) \geq P(X\leq X_0-t)$, prove that $E(X)\geq X_0$.

Intuitively, it is cleared to me that the expectation is no less than $X_0$ as the distribution is skewed to the right (the chance of $X$ to rise at least $t$ is no less than the chance of $X$ drops at least $t$). But I couldn't prove it rigorously.

Here is what i tried so far: Let $f(x)=F'(x)$ be the density function. It is easy to deduce that $\int_{X_0}^{\infty} f(x) dx\geq \frac{1}{2}$ and $\int_0^{X_0} f(x) dx \leq \frac{1}{2}.$

Then $$E(X)=\int_0^{\infty}xf(x)dx=\int_0^{X_0}xdF(x) + \int_{X_0}^{\infty} xf(x) dx$$
$$=[xF(x)]_0^{X_0}-\int_0^{X_0}F(x)dx + \int_{X_0}^{\infty} xf(x) dx$$
$$=X_0F(X_0)-\int_0^{X_0}\int_0^x f(t) dt dx + \int_{X_0}^{\infty} xf(x) dx$$
$$\geq X_0F(X_0)-\int_0^{X_0}\int_0^{X_0} f(t) dt dx + \int_{X_0}^{\infty} X_0f(x) dx$$

$$\geq X_0F(X_0) -\int_0^{X_0} \frac{1}{2} dx + \frac{X_0}{2}$$
$$=X_0F(X_0) -\frac{X_0}{2} + \frac{X_0}{2}=X_0F(X_0)$$

But then $F(X_0) \leq 1$. Please help