An elegant proof for a claim about orthogonal and positive-definite matrices

The following proof does diagonalize $P$, but not in matrix form.

Let $\{v_1,\ldots,v_n\}$ be an orthonormal eigenbasis of $P$ and $Pv_i=\lambda_iv_i$ for each $i$. Then $$ \langle P,I\rangle = \langle P,O\rangle = \sum_i\langle Pv_i,Ov_i\rangle \le \sum_i\|Pv_i\|\|Ov_i\| = \sum_i\lambda_i = \operatorname{tr}(P) = \langle P,I\rangle. $$ Therefore $\langle Pv_i,Ov_i\rangle=\|Pv_i\|\|Ov_i\|$ for each $i$. Since $Pv_i=\lambda v_i$ and $Ov_i$ is a unit vector, we have $\langle v_i,Ov_i\rangle=1$ and in turn $Ov_i=v_i$ whenever $\lambda_i>0$. Hence $OP$ and $P$ agree on $\{v_1,\ldots,v_n\}$, meaning that $OP=P$.

Trace is invariant under cyclic permutations, so $$\mathrm{tr}(O^TP)=\mathrm{tr}(P^{1/2}O^TP^{1/2})=\langle OP^{1/2},P^{1/2}\rangle.$$ Letting $\|\cdot\|$ denote the Frobenius norm, one form of the Cauchy-Schwarz inequality says that

$$\langle O,P\rangle=\langle OP^{1/2},P^{1/2}\rangle\leq \|OP^{1/2}\|\cdot\|P^{1/2}\|=\mathrm{tr}(P)^{1/2}\mathrm{tr}(P)^{1/2}=\langle I,P\rangle$$ with equality only when

$$OP^{1/2}=P^{1/2}\frac{\|OP^{1/2}\|}{\|P^{1/2}\|}=P^{1/2}.$$ Multiplying on the right by $P^{1/2},$ this condition implies $$OP=P.$$