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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.

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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 $\|\vec y||^{2}=\vec y^{T} \vec y$.

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Surprisingly, I got the answer from my graduate analysis course in the Hilbert spaces. The hint given by @Kabo Murphy is also well stated. For future readers, it is the Bessel's inequality.

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