Show, that $\frac {E[Xg(X)]}{E[g(X)]} \ge E[X]$ when $g$ strictly monotonic increasing

Let $$g:\mathbb R \to (0,\infty), X$$ real valued random variable and $$g(X) \in \mathcal L^2$$ and $$g$$ strictly monotonic increasing.

Show, that $$\frac {E[Xg(X)]}{E[g(X)]} \ge E[X]$$

I tried something with expected values and their correlation with covariance, but I don't get the final result.

• Strict inequality doesn't hold since $X$ could have been almost surely constant. Jun 14 '19 at 15:10
• How would you argue for $\ge$ @user10354138 Jun 14 '19 at 15:24
• Is this homework? Exam? If the former, I can give a hint. If the latter, I can give a maybe tiny tiny hint...? Jun 14 '19 at 15:30
• @antkam It's from last years exam. Jun 14 '19 at 15:34

Since (the identity and) $$g$$ is strictly increasing, the covariance of $$X$$ and $$g(X)$$ is nonnegative. Hence $$0\leq \operatorname{Cov}(X,g(X))= \mathbb{E}[Xg(X)]-\mathbb{E}[g(X)]\mathbb{E}X$$ and we can divide by $$\mathbb{E}[g(X)]>0$$.
Hint: Let $$f$$ and $$g$$ both be monotonically increasing functions. Let $$X_1, X_2$$ be i.i.d. copies of $$X$$ and consider the sign of $$(g(X_1)-g(X_2))(f(X_1)-f(X_2)).$$