Let $E_1, E_2$ be intermediate fields of the extension $K\subseteq F$. Then $\mathrm{tr.d.} (E_1E_2/K) \geq \mathrm{tr.d.} (E_i/K)$ for $i=1,2$.

Question

I want to prove this: Let $E_1, E_2$ be intermediate fields of the extension $K\subseteq F$. Then we have followings:

1. $\mathrm{tr.d.} (E_1E_2/K) \geq \mathrm{tr.d.} (E_i/K)$ for $i=1,2$;
2. $\mathrm{tr.d.} (E_1E_2/K) \leq \mathrm{tr.d.}(E_1/K) + \mathrm{tr.d.}(E_2/K)$.

Definition: Let $F$ be an extension field of $K$. A transcendence base of $F$ over $K$ is a subset $S$ of $F$ which is algebraically independent over $K$ and is maximal (with respect to set-theoretic inclusion) in the set of all algebraically independent subsets of $F$.

Definition: Let $F$ be an extension field of $K.$ The transcendence degree of $F$ over $K$ (denoted $\mathrm{tr.d.}F/K$) is the cardinal number $|S|$, where $S$ is any transcendence base of $F$ over $K.$

(Related) theorem: If $F$ is an extension field of $E$ and $E$ an extension field of $K,$ then \begin{equation*} \mathrm{tr.d.}F/K=(\mathrm{tr.d.}F/E)+(\mathrm{tr.d.}E/K). \end{equation*}

Does anyone have idea how to prove 1 and 2? Thank you.