How to show that the square root of a matrix is concave?

Question

Let $A$ and $B$ be two symmetric positive semidefinite matrices. How can we prove that $$\sqrt{\lambda A + (1 - \lambda) B} \succeq \lambda \sqrt{A} + (1 - \lambda) \sqrt{B}$$ for any $\lambda \in (0, 1)$? Here, $\sqrt{A}$ represents the square root of $A$. And the notation $A \succeq B$ represents that $A - B$ is positive semidefinite.

Answer 1

Start with $(\sqrt{A}-\sqrt{B})^2\geq0$. Expanded, this is $$ \sqrt{A}\sqrt{B}+\sqrt{B}\sqrt{A}\leq A+B. $$ Then $$ (\lambda-\lambda^2)(\sqrt{A}\sqrt{B}+\sqrt{B}\sqrt{A})\leq(\lambda-\lambda^2)( A+B). $$ Since $\lambda-\lambda^2=(1-\lambda)-(1-\lambda)^2$, $$ (\lambda-\lambda^2)(\sqrt{A}\sqrt{B}+\sqrt{B}\sqrt{A})\leq(\lambda-\lambda^2) A+[(1-\lambda)-(1-\lambda)^2]B, $$ which we can rewrite as $$ \lambda^2A+(1-\lambda)^2B+(\lambda-\lambda^2)(\sqrt{A}\sqrt{B}+\sqrt{B}\sqrt{A})\leq\lambda A+(1-\lambda)B. $$ That is, $$ (\lambda\sqrt A+(1-\lambda)\sqrt B)^2\leq \lambda A+(1-\lambda)B. $$ As the square root is operator monotone, $$ \lambda\sqrt A+(1-\lambda)\sqrt B\leq \sqrt{\lambda A+(1-\lambda)B}. $$