I am trying to prove the following inequality:
$$ x^T(I_d-11^T/d)x \ge 0 $$
where $d$ is a non-negative integer, $I_d$ a $d\times d$ identity matrix and $1$ a $d \times 1$ vector composed of ones. I can prove this by using the lemma that for any real symmetric $d \times d$ matrix $A$ the following holds:
$$ \lambda_{(1)} \le \frac{x^TAx}{x^Tx} \le \lambda_{(d)} $$
where $\lambda_{(1)}$ is the smallest and $\lambda_{(d)}$ the largest eigenvalue. Combining this in the above equation one can show (if I am not mistaken) that
$$ 0 \le x^T(I_d-11^T/d)x \le x^Tx $$
which proves the semi-positive definiteness.
Are there any more intuitive ways to prove this without the appeal to this lemma?
Sure. Let $v$ denote the unit-vector $1/\sqrt{d}$. Then $$ (I - 11^T/d) = (I - vv^T) $$ It follows that $$ x^T(I - 11^T/d)x = x^T(I - vv^T)x^T = x^Tx - (x^Tv)^2 =\\ \|x\|^2 - |\langle x,v \rangle|^2 $$ and by Cauchy-Schwarz, this must be non-negative.