Proof of the inequality $\sqrt{\det X} \leq \frac{\operatorname{tr}X}{2}$ 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". )


 A: 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 $a_1=a_2$.
A: If I understand correctly I think you are trying to prove:
$$\sqrt{\det(h_{j}^{i})}\le\frac{1}{2}\text{trace}(h_{j}^{i})$$
where $(h_{j}^{i})$ is a positive definite $2\times2$ matrix. This follows from the fact that the determinant is the product of the eigenvalues, the trace is the sum of the eigenvalues, and the Arithmetric Geometric Inequality. The case of equality also follows from the Arithmetric Geometric Inequality.
