Problem:
A is an $n\times n$ complex matrix. Let $\lambda _\max(Y)$ and $\sigma _\max(Y)$ denote respectively the largest eigenvalue and the largest singular value of a square matrix $Y$. The question is to prove that: $\lambda _\max\left(\frac{A+A^*}{2}\right)\leq \sigma _\max (A)$
I started as follows: Since $A$ is Hermitian, then $(A+A^*)$ is Hermitian as well. Then: $\lambda_\max\left(\frac{A+A^*}{2}\right)=\max_{\left \| x \right \|=1}x^*\left(\frac{A+A^*}{2}\right)x$ where $x\in \mathbb{C}^n$
I need to prove the following to finish my proof: $\max_{\left \| x \right \|=1}x^*\left(\frac{A+A^*}{2}\right)x\geq \max_{\left \| x \right \|=1}(x^*A^*Ax)^{\frac{1}{2}}\tag{1}$
From the first sight, it looks like I have to use the fact that: $\sqrt{ab}\geq \frac{a+b}{2}$ for $a$ and $b$ positive, but this works only for numbers and not matrices.
I appreciate if someone can prove the inequality $(1)$.