I have a answer to a question about trace. Is there an easier answer to this question? Let $A\in M_n(\mathbb{C})$. Show that
$$tr\left(\frac{A+A^*}{2}\right)\leq tr((A^*A)^{1/2}).$$
My answer:
It is easy to see that
$$tr\left(\frac{A+A^*}{2}\right)=\text{Re}(tr(A))\qquad and\qquad tr((A^*A)^{1/2})=\sum_{i=1}^n\sigma_i(A),$$
where $\sigma(A)$ is the singular value of $A$.
On the other hand, based on Polar decomposition theorem, there exists a unitary matrix $P$ such that $A=P((A^*A)^{1/2})$.
Now, according to [Theorem 8.7.6, Page 551, R. Horn, C. Johnson, Matrix analysis, Second edition], we have:
$$\begin{eqnarray}
 \text{Re}(tr(A))&=\text{Re}(tr(P((A^*A)^{1/2})))\\
&\leq\sum_{i=1}^n \sigma_i(P)\sigma_i ((A^*A)^{1/2}).
 \end{eqnarray} $$
Since $P$ is unitary, $\sigma_i(P)=1$. We have also $\sigma_i ((A^*A)^{1/2})=\sigma_i(A)$. So,
$$ \text{Re}(tr(A))\leq\sum_{i=1}^n \sigma_i(A).$$
Therefore, we get
$$tr\left(\frac{A+A^*}{2}\right)\leq tr((A^*A)^{1/2}).$$

Now,  is there an easier answer to this question?
 A: You’ve presented a nice clean proof of that trace inequality, and one could hardly expect to find a shorter argument.  The proof of Theorem 8.7.6 given in Horn and Johnson does, however, rely on some results for doubly stochastic matrices (such as Birkhoff’s theorem) that are not exactly obvious.
In the first edition of Matrix Analysis, though, Horn and Johnson have a fairly straightforward exercise that provides an even stronger result than the trace inequality that you presented.  Namely, if $A$ is a complex $n\times n$ matrix, $\lambda_1 \ge \ldots \ge \lambda_n$ are the ordered eigenvalues of  $\frac{1}{2}\left[{A + A^*} \right]$, and $\sigma_1 \ge \ldots \ge \sigma_n$ are the ordered singular values of $A$, then $\lambda_i \left( {\frac{1}{2}\left[ {A + A^*} \right]} \right) \le \sigma_i \left( A \right)$  for $1 \le i \le n.$  This result, which immediately implies the trace inequality, is outlined in exercise 15 on page 454 (section 7.4) of the first edition of Matrix Analysis.
The outline consists of simply showing that $$\frac{1}{2}y^* \left( {A + A^* } \right)y = {\mathop{\rm Re}\nolimits} (y^* Ay) \le \left\| {Ay} \right\|_2$$ for a Euclidean unit vector $y$, and then applying the Courant-Fischer theorem to $\frac{1}{2}\left[{A + A^*}\right]$ on the left side of the inequality, and the singular-value analog of the Courant-Fischer theorem to the right side of the inequality.  If not easier, it is at least arguably more elementary (and stronger) than the proof using Theorem 8.7.6 from Horn and Johnson's book.
