It is easy to prove this by taking point $(a\cos\theta, b\sin\theta)$ on $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$. But I am looking for purely geometric proof without using trigonometry or coordinates.
It tried it by using triangle inequality: I took an arbitrary point $P$ on the ellipse with foci $S_1$, $S_2$ and centre $C$. Then we get (by using triangle inequalities two times) $$ CP < a + ae, $$ where $a$ is major axis and $e$ eccentricity. But it is not exactly that I want. If somehow I could show $CP < a$, then I would be done.
Please help.
Let $PQ$ be any chord of the ellipse. From triangle inequality we have: $$ S_1P+S_1Q\ge PQ\quad\text{and}\quad S_2P+S_2Q\ge PQ, $$ where equality holds only if $PQ$ passes through $S_1$ or $S_2$. If $PQ$ doesn't pass through both foci we have then: $$ S_1P+S_2P+S_1Q+S_2Q> 2PQ,\quad\text{that is:}\quad 2a> PQ, $$ Q.E.D.