Dimension of isometry group of complete connected Riemannian manifold

Given an $n$-dimensional geodesically complete connected Riemannian manifold $M$, we want to prove that the dimension of its isometry group is $$\dim {\rm ISO}(M) \leq \frac{n(n+1)}2.$$

Does it suffice to say that, since Euclidean space $\mathbb{R}^n$ is expected to be maximally symmetric, and the number above is the dimension of its isometry group, namely translations plus rotations, then the bound must be true for any other manifold? Do you know a more rigorous proof?

The comparison with Euclidean space gives the correct intuition but to give further details one could mention that the orbit of a point $p\in M$ under the isometry group is a homogeneous space of the isometry group, whose dimension is at most that of $M$ itself, namely $n$. Now the action of the stabilizer at $p\in M$ is determined uniquely by the induced action on the tangent space at $p$, which can be identified with a subgroup of $SO(n)$. The latter is known to have dimension $n(n-1)/2$, so for the total dimension of the isometry group you get at most $n + n(n-1)/2 = n(n+1)/2$.