= Solution
Let $p=\sum_{j=1}^m\langle u,v_j\rangle v_j$. <Orthonormality> gives $\langle u-p,v_k\rangle=0$ for every $k$, so $u-p$ is <orthogonal> to $p$. The <Pythagorean theorem for inner product spaces> gives
$$
\|u\|^2=\|u-p\|^2+\|p\|^2
=\|u-p\|^2+\sum_{j=1}^m\langle u,v_j\rangle^2.
$$
Since a squared <norm> is nonnegative,
$$
\boxed{\langle u,u\rangle\geq\sum_{j=1}^m\langle u,v_j\rangle^2.}
$$
This is <Bessel's inequality>, with equality exactly when $u$ belongs to the <span> of the <orthonormal set>.
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