It is well-known as Young inequality that for $0 \leq \nu \leq 1$ and $a, b \geq 0$ $$ \nu a + (1 - \nu)b \geq a^\nu b^{1-\nu} \geq \frac{a + b - |a- b|}{2}. $$

When $n = 2$ and $\nu = \frac{1}{2}$, the inequality $a^\frac{1}{2}b^\frac{1}{2} \geq \frac{a+b - |a-b|}{2}$ is called the reverse Caucy inequaality. A natural matrix form for two positive definite matrices $A$ and $B$ could be written as $$ A^\frac{1}{2}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^\frac{1}{2}A^\frac{1}{2} \geq \frac{A + B}{2} - \frac{|A - B|}{2}. $$ Furuichi, however, pointed out that the trace inequality $$ Tr(A^\frac{1}{2}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^\frac{1}{2}A^\frac{1}{2}) \geq \frac{1}{2}Tr(A + B - |A - B|) $$ is not true in general. Note that $A\sharp B = A^\frac{1}{2}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^\frac{1}{2}A^\frac{1}{2}$ is the geometric mean and $A\nabla_\nu B = \nu A + (1 - \nu)B$ is the weighted arithmetric mean of $A$ and $B$, respctively.

In this talk we introduce the operator inequalities for operator means such that $$ \phi(A) \sigma \phi(B) \geq \phi(A \nabla_\nu B) $$ for a non-negative operator convex function $\phi$ on $[0, \infty)$ and all positive definite matrices $A$ and $B$, and we show the characterization of $\sigma$.