TITLE: Proof of the inequality $\sqrt{\det X} \leq \frac{\operatorname{tr}X}{2}$
QUESTION [7 upvotes]: Let $A, B \in M_2(\mathbb{R})$ be symmetric and positive definite. Put $X:=AB$. 
then, we have the following inequality:
$$\sqrt{\det X}\leq \dfrac{1}{2}\operatorname{trace}X.$$
and the equality holds iff $\exists \lambda>0\ s.t.\ X=\lambda E$. 
I cannot prove this though I have put 
$A=\left(\begin{array}{ccc}
a &s\\
s &b\\
\end{array} 
\right), $
$B=\left(\begin{array}{ccc}
x &t\\
t &y\\
\end{array} 
\right), $
 and have extended each part of the inequality by $a,s,b,x,t,y$. Please help me. 
(I encountered this problem when I tried to solve [3.10] of Nishikawa "Variational Problems in Geometry". )

REPLY [4 votes]: Since $A\geq 0$, there is a unique $A^{1/2}\geq 0$ such that $(A^{1/2})^2=A$. Observe furthermore that 
$$
\mbox{det} (AB)=\mbox{det} (A)\mbox{det} (B)=\mbox{det} (A^{1/2})\mbox{det} (B)\mbox{det} (A^{1/2})= \mbox{det} (A^{1/2}BA^{1/2}),
$$
and $\mbox{tr}(AB)=\mbox{tr}(A^{1/2}A^{1/2}B)=\mbox{tr}(A^{1/2}BA^{1/2})$. Since $A^{1/2}BA^{1/2}\geq 0$, it suffices to show that
$$
\sqrt{\mbox{det} A}\leq \frac{1}{2}\mbox{tr} A
$$
when $A\geq 0$. Let $a_1,a_2\geq 0$ be the eigenvalues of $A$, then
$$
\sqrt{\mbox{det} A}=\sqrt{a_1 a_2}\leq \frac{a_1+a_2}{2}=\frac{1}{2}\mbox{tr}A,
$$
by Young's inequality for $p=q=2$. Equality occurs in Young's inequality if and only if $a_1=a_2$.