An inequality relates to a set of pairwise orthogonal unit vectors

Question:

Let $$\vec{u}_1, \dots, \vec{u}_k$$ be pairwise orthogonal unit vectors in $$\mathbb{R}^n$$. Prove that $$\vec{c}^\mathsf{T} \vec{c} \geq \sum_{i=1}^k(\vec{c}^\mathsf{T} \vec{u}_i)^2$$ for all $$\vec{c} \in \mathbb{R}^n$$.

My attempt is the following:

Define $$\vec{c} = [c_1 \dots c_n]^\mathsf{T}$$and rewrite $$\vec{c}^\mathsf{T} \vec{c} = \sum_{i=1}^n c_i^2$$ Also, if $$\vec{u}_1, \dots, \vec{u}_k$$ are pairwise orthogonal unit vectors in $$\mathbb{R}^n$$, we have $$k \leq n$$ (this seems to be true, but I don't know the reference). Therefore, to show the above inequality, it suffices to show that $$c_i \geq \vec{c}^\mathsf{T} \vec{u}_i$$ for all $$i = 1, \dots, k$$. I don't what should do next and how to apply the property of pairwise orthogonal unit vectors. Any help is appreciated.

Hint: Verify that $$\vec c^{T}\vec c-\sum\limits_{k=1}^{n}(\vec c^{T} \vec u_i)^{2}=\|\vec c-\sum\limits_{k=1}^{n} \vec (c^{T} \vec u_i) u_i\|^{2}$$ using the fact that $$\|\vec y||^{2}=\vec y^{T} \vec y$$.