Throughout this paper, a capital letter means a bounded linear operator on Hilbert space $H$. An operator $A$ is called positive, in symbol, $A\geq0$ if $(Ax, x)\geq 0$ for all $x\in H$. $T$ is called strictly positive (simply $T > 0$) if $T$ is positive and invertible, $\alpha_{i}$ are positive numbers with $\sum\limits_{i=1}\limits^{n}\alpha_{i}=1$.

A continuous function $f$ on interval $I$ is called operator convex on $I$, if $\sigma(A)$, $\sigma(B)\subset I$,

$f((1-\alpha)A+\alpha B)\leq(1-\alpha)f(A)+\alpha f(B)\quad\mbox{holds for}\quad\alpha\in[0, 1]. $

(1.1)

Convex functions and operator convex functions are different. Typical example of such function is $t^{r}$ on $(0, \infty)$, which is a convex function for $r>2$, but is not operator convex.

In [1], Ando, Li and Mathias proposed a definition of the geometric mean for an $n$-tuple of positive operators and showed that it has many required properties on the geometric mean.Following [3, 4], we recall the definition of the weighted geometric mean $G[n,t]$ with $t\in [0,1]$ for an $n$-tuple of positive invertible operators $A_{1},A_{2},\cdots, A_{n}$. Let $G[2, t](A_{1}, A_{2})=A_{1}\sharp_{t} A_{2}$. For $n\geq 3$, $G[n,t]$ is defined inductively as follows: put ${A_{i}^{(0)}}=A_{i}$ for all $i=1, 2, \cdots, n$, and

$A_{i}^{(r)}=G[n-1, t]((A_{j}^{(r-1)})_{j\neq i})=G[n-1, t](A_{1}^{(r-1)}, \cdots, A_{i-1}^{(r-1)}, A_{i+1}^{(r-1)}, \cdots, A_{n}^{(r-1)}) $

inductively for $r$. Then the sequence $\{A_{i}^{(r)}\}_{r=0}^{\infty}$ have the same limit for all $i=1, 2, \cdots, n$ in the Thompson metric. So we can define

$G[n, t](A_{1}, A_{2}, \cdots, A_{n})=\lim\limits_{r\rightarrow \infty} A_{i}^{(r)}. $

Similarly, we can define the weighted arithmetic mean as follows: Let $A[2, t](A_{1}, A_{2})=(1-t)A_{1}+tA_{2}.$ For $n\geq 3$, put $\tilde{A}_{i}^{(0)}=A_{i}$ for all $i=1, 2, \cdots, n$ and

$\tilde{A}_{i}^{(r)}=A[n-1, t]((\tilde{A}_{j}^{(r-1)})_ {j\neq i})=A[n-1, t](\tilde{A}_{1}^{(r-1)}, \cdots, \tilde{A}_{i-1}^{(r-1)}, \tilde{A}_{i+1}^{(r-1)}, \cdots, \tilde{A}_{n}^{(r-1)}). $

The sequence $\{\tilde{A}_{i}^{(r)}\}$ have the same limit for all $i=1, 2, \cdots, n$, so it's expressed by

$A[n, t](A_{1}, A_{2}, \cdots, A_{n})=\lim\limits_{r\rightarrow \infty} \tilde{A}_{i}^{(r)}. $

Here we introduce the following power means: for positive invertible operators $A_{1},A_{2},\cdots, A_{n}$, define

$\begin{eqnarray*} F(r)=\begin{cases}(\nabla_{\alpha}(A_{1}^{r}, A_{2}^{r}, \cdots, A_{n}^{r}))^{\frac{1}{r}}, &r\neq 0; \cr \exp(\nabla_{\alpha}(\log A_{1}, \log A_{2}, \cdots, \log A_{n})), &r=0.\end{cases}\end{eqnarray*}$

It is clear that $F(r)$ is monotone increasing under the chaotic order, but is not monotone under the usual order. Besides, $F(0)$ is called chaotically geometric mean for $A_{1},A_{2},\cdots, A_{n}$.

Theorem A [6-8] Let $0<m\leq A_{i}\leq M$ with $m<M$ for $i=1, 2, \cdots, n$ and $ \sum\limits_{i=1}\limits^{n}\parallel x_{i}\parallel^{2}=1$. If $f(t)$ is a positive real valued continuous convex function on $[m, M]$, then

$f(\sum\limits_{i=1}\limits^{n}(A_{i}x_{i}, x_{i}))\leq \sum\limits_{i=1}\limits^{n}(f(A_{i})x_{i}, x_{i})\leq \lambda(m, M, f)f(\sum\limits_{i=1}\limits^{n}(A_{i}x_{i}, x_{i})), $

(1.2)

where

$\lambda(m, M, f)=\max\{\frac{1}{f(t)}[\frac{f(M)-f(m)}{M-m}(t-m)+f(m)];\ \ t\in[m, M]\}. $

(1.3)

Theorem B [4] Let $A$ and $B$ be positive operators satisfying $0<m\leq A\leq M$ for some scalars $m<M$. If $0\leq A\leq B$, then

$A^{p}\leq K_+(h, p)B^{p},\ \ p\geq1, $

where

$K_{+}(h, p)=\frac{(p-1)^{p-1}}{p^{p}}\frac{(h^{p}-1)^{p}}{(h-1)(h^{p}-1)^{p-1}}, \quad h=\frac{M}{m}. $

(1.4)

Theorem C [3] If $0<m\leq A_{i}\leq M $ with $m<M$, $i=2, \cdot\cdot\cdot, n$ ($n\geq 2$), then

$G[n, t](A_{1}, A_{2}, \cdots, A_{n})\leq A[n, t](A_{1}, A_{2}, \cdots, A_{n})\leq K(h, 2)G[n, t](A_{1}, A_{2}, \cdots, A_{n}), $

where

$K(h, 2)=\frac{(h+1)^{2}}{4h},\quad h=\frac{M}{m}. $

Based on the previous results coming from the method we obtain the ratio type inequalities as follows.

Theorem 2.1 Let $0<m\leq A_{i}\leq M$ with $m<M$ for $i=1, 2, \cdots, n$. If $f(t)$ is a positive real valued continuous convex function on $[m, M]$, then

$\begin{eqnarray*}\frac{1}{\lambda(m, M, f)}f(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n}))&\leq& \nabla_{\alpha}(f(A_{1}), f(A_{2}), \cdots, f(A_{n}))\\ &\leq& \lambda(m, M, f)f(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n})). \end{eqnarray*}$

Proof For unit vector $x\in H$, put $x_{i}=\sqrt{\alpha_{i}}x, i=1, 2, \cdots, n$ in Theorem A, then

$\sum\limits_{i=1}\limits^{n}\alpha_{i}(f(A_{i})x, x)\leq \lambda(m, M, f)f(\sum\limits_{i=1}\limits^{n}\alpha_{i}(A_{i}x, x)).$

Since $f(t)$ is a positive real valued continuous convex function on $[m, M]$, (1.2) leads to

$\begin{eqnarray*} ((\sum\limits_{i=1}\limits^{n}\alpha_{i}f(A_{i}))x, x) &\leq& \lambda(m, M, f)f((\sum\limits_{i=1}\limits^{n}\alpha_{i}A_{i}x, x))\leq \lambda(m, M, f)(f(\sum\limits_{i=1}\limits^{n}\alpha_{i}A_{i})x, x).\end{eqnarray*}$

Thus we have

$\nabla_{\alpha}(f(A_{1}), f(A_{2}), \cdots, f(A_{n}))\leq \lambda(m, M, f)f(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n})).$

Next, since $0<m\leq\sum\limits_{i=1}\limits^{n}\alpha_{i}A_{i}\leq M$ and $f(t)$ be a positive real valued continuous convex function on $[m, M]$, it follows from (1.2) that

$\begin{eqnarray*}(\nabla_{\alpha}(f(A_{1}), f(A_{2}), \cdots, f(A_{n}))x, x)&=&\sum\limits_{i=1}\limits^{n}\alpha_{i}(f(A_{i})x, x) \geq f(\sum\limits_{i=1}\limits^{n}\alpha_{i}(A_{i}x, x)) \\&=&f(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n})x, x)\\ &\geq&\frac{1}{\lambda(m, M, f)}(f(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n}))x, x).\end{eqnarray*}$

Therefore we have

$\lambda(m, M, f)\nabla_{\alpha}(f(A_{1}), f(A_{2}), \cdots, f(A_{n}))\geq f(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n})).$

By the same way we can get the counterpart of Theorem 2.1.

Theorem 2.2 Let $0<m\leq A_{i}\leq M$ with $m<M$ for $i=1, 2, \cdots, n$. If $f(t)$ is a positive real valued continuous concave function on $[m, M]$, then

$\begin{eqnarray*}\frac{1}{\mu(m, M, f)}f(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n}))&\geq& \nabla_{\alpha}(f(A_{1}), f(A_{2}), \cdots, f(A_{n})) \\ &\geq& \mu(m, M, f)f(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n})), \end{eqnarray*}$

where $\mu(m, M, f)=\min\{\frac{1}{f(t)}(\frac{f(M)-f(m)}{M-m}(t-m)+f(m)), \ \ t\in[m, M]\}. $

Theorem 3.1  Let $0<m\leq A_{i}\leq M$ with $m<M$ for $i=1, 2, \cdots, n$.

(ⅰ) If $0<r\leq1$, then

$(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n}))^{r}\geq \nabla_{\alpha}(A_{1}^{r}, A_{2}^{r}, \cdots, A_{n}^{r})\geq K_{+}(h^{r}, \frac{1}{r})^{-r}(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n}))^{r};$

(ⅱ) If $1\leq r\leq2$, then

$(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n}))^{r}\leq \nabla_{\alpha}(A_{1}^{r}, A_{2}^{r}, \cdots, A_{n}^{r})\leq K_{+}(h, r)(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n}))^{r};$

(ⅲ) If $r>2$, then

$\frac{1}{K_{+}(h, r)}(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n}))^{r}\leq \nabla_{\alpha}(A_{1}^{r}, A_{2}^{r}, \cdots, A_{n}^{r})\leq K_{+}(h, r)(\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n}))^{r}. $

Proof Put $f(t)=t^{r}$, we distinguish three cases.

In the case of $0<r\leq1$, $f(t)=t^{r}$ is operator concave, $\mu(m, M, f)=K_{+}(h^{r}, \frac{1}{r})^{-r}$, Theorem 2.2 and the definition of operator concavity of $t^{r}$ lead to (ⅰ).

In the case of $1\leq r\leq2$, $f(t)=t^{r}$ is operator convex, $\lambda(m, M, f)=K_{+}(h, r)$, following from Theorem 2.1 and the definition of operator convex, we have (ⅱ).

In the case of $r>2$, $f(t)=t^{r}$ is not operator convex, but is convex, so Theorem 2.1 yields (ⅲ).

Theorem 3.2 Let $0<m\leq A_{i}\leq M$ with $m<M$ for $i=1, 2, \cdots, n$, and $0<r\leq s$.

(ⅰ) If $0<r\leq1$, then

$K_{+}(h^{r}, \frac{1}{r})^{-1}K_{+}(h^{r}, \frac{s}{r})^{-\frac{1}{s}}F(s)\leq F(r)\leq K_{+}(h^{r}, \frac{1}{r})F(s);$

(ⅱ) If $r\geq1$, then

$K_{+}(h^{r}, \frac{s}{r})^{-\frac{1}{s}}F(s)\leq F(r)\leq F(s). $

Proof Since $0<\frac{r}{s}\leq1$, $m^{s}\leq A^{s}_{i}\leq M^{s}$, it follows from Theorem 3.1 that

$\begin{eqnarray}(\nabla_{\alpha}(A_{1}^{s}, A_{2}^{s}, \cdots, A_{n}^{s}))^{\frac{r}{s}}&\geq& (\nabla_{\alpha}(A_{1}^{r}, A_{2}^{r}, \cdots, A_{n}^{r})) \nonumber\\ &\geq& K_{+}(h^{r}, \frac{s}{r})^{-\frac{r}{s}}(\nabla_{\alpha}(A_{1}^{s}, A_{2}^{s}, \cdots, A_{n}^{s}))^{\frac{r}{s}}. \end{eqnarray}$

(3.1)

If $r\geq1$, then $0<\frac{1}{r}\leq1$, by raising all terms of (3.1) to $\frac{1}{r}$ it follows from operator monotonity of $x^{\frac{1}{r}}$ that

$(\nabla_{\alpha}(A_{1}^{s}, A_{2}^{s}, \cdots, A_{n}^{s}))^{\frac{1}{s}}\geq (\nabla_{\alpha}(A_{1}^{r}, A_{2}^{r}, \cdots, A_{n}^{r}))^{\frac{1}{r}}\geq K_{+}(h^{r}, \frac{s}{r})^{-\frac{1}{s}}(\nabla_{\alpha}(A_{1}^{s}, A_{2}^{s}, \cdots, A_{n}^{s}))^{\frac{1}{s}}.$

If $0<r\leq1$, then $\frac{1}{r}\geq1$, Theorem B and (3.1) lead to

$\begin{eqnarray*}K_{+}(h^{r}, \frac{1}{r})(\nabla_{\alpha}(A_{1}^{s}, A_{2}^{s}, \cdots, A_{n}^{s}))^{\frac{1}{s}} &\geq& \nabla_{\alpha}(A_{1}^{r}, A_{2}^{r}, \cdots, A_{n}^{r})^{\frac{1}{r}}\\ &\geq& K_{+}(h^{r}, \frac{1}{r})^{-1}K_{+}(h^{r}, \frac{s}{r})^{-\frac{1}{s}}(\nabla_{\alpha}(A_{1}^{s}, A_{2}^{s}, \cdots, A_{n}^{s}))^{\frac{1}{s}}. \end{eqnarray*}$

Theorem 3.3 If $0<m\leq A_{i}\leq M$ with $m<M$ for $i=1, 2, \cdots, n$, then

$\frac{1}{M_{h}(1)}e^{\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n})}\leq \nabla_{\alpha}(e^{A_{1}}, e^{A_{2}}, \cdots, e^{A_{n}})\leq M_{h}(1)e^{\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n})},$

where $M_{h}(1)=\frac{h^{\frac{1}{h-1}}}{e\log h^{\frac{1}{h-1}}}$, $h=e^{M-m}.$

Proof Put $f(t)=e^{t}$ in Theorem 2.1, then the required inequalities hold since

$\begin{eqnarray*} \lambda(m, M, e^{t})&=&\frac{1}{e^{t_{0}}}(\frac{e^{M}-e^{m}}{M-m}(t_{0}-m)+e^{m}), \quad \mbox{where}\quad t_{0}=\frac{(m+1)e^{M}-(M+1)e^{m}}{e^{M}-e^{m}}\\ &=&\frac{e^{M}-e^{m}}{M-m}e^{-\frac{(m+1)e^{M}-(M+1)e^{m}}{e^{M}-e^{m}}} =\frac{h-1}{\log h}e^{-1+\frac{M-m}{h-1}} =\frac{h^{\frac{1}{h-1}}}{e\log h^{\frac{1}{h-1}}} = M_{h}(1). \end{eqnarray*}$

Theorem 3.4 If $0<m\leq A_{i}\leq M$ with $m<M$ for $i=1, 2, \cdots, n$, then

$\frac{1}{M_{h}(1)}\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n})\leq \diamondsuit_{\alpha}(A_{1}, A_{2}, \cdots, A_{n})\leq M_{h}(1)\nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n}),$

where $h=\frac{M}{m}$.

Proof Replace $A_{i}$ by $\log A_{i}$ in Theorem 3.3, then $h=\exp({\log M-\log m})=\frac{M}{m}$, and

$\begin{eqnarray*}&& \frac{1}{M_{h}(1)}\exp({\sum\limits_{i=1}\limits^{n}\alpha_{i}\log A_{i}})\leq \sum\limits_{i=1}^{n}\alpha_{i}A_{i}\leq M_{h}(1)\exp({\sum\limits_{i=1}\limits^{n}\alpha_{i}\log A_{i})},\\ && \frac{1}{M_{h}(1)}\diamondsuit_{\alpha}(A_{1}, A_{2}, \cdots, A_{n})\leq \nabla_{\alpha}(A_{1}, A_{2}, \cdots, A_{n})\leq M_{h}(1)\diamondsuit_{\alpha}(A_{1}, A_{2}, \cdots, A_{n}).\end{eqnarray*}$

Theorem 3.5 Let $0<m\leq A_{i}\leq M$ with $m<M$ for $i=1, 2, \cdots, n$, and $0\leq t\leq 1$.

(ⅰ) If $0<r\leq 1$, then

$\begin{eqnarray*}K_{+}(h^{r}, \frac{1}{r})^{-r}K(h^{r}, 2)^{-1}(G_{[n, t]}(A_{1}, \cdot\cdot\cdot, A_{n}))^{r} &\leq& G_{[n, t]}(A_{1}^{r}, \cdot\cdot\cdot, A_{n}^{r})\\ &\leq& K(h, 2)^{r} (G_{[n, t]}(A_{1}, \cdot\cdot\cdot, A_{n}))^{r}.\end{eqnarray*}$

(ⅱ) If $1\leq r\leq2$, then

$\begin{eqnarray*}K_{+}(h, r)^{-1}K(h^{r}, 2)^{-1}(G_{[n, t]}(A_{1}, \cdot\cdot\cdot, A_{n}))^{r} &\leq& G_{[n, t]}(A_{1}^{r}, \cdot\cdot\cdot, A_{n}^{r}) \\ &\leq& K_{+}(h, r)^{2}K(h, 2)^{r} (G_{[n, t]}(A_{1}, \cdot\cdot\cdot, A_{n}))^{r}.\end{eqnarray*}$

(ⅲ) If $r>2$, then

$\begin{eqnarray*}K_{+}(h, r)^{-2}K(h^{r}, 2)^{-1}(G_{[n, t]}(A_{1}, \cdots, A_{n}))^{r} &\leq& G_{[n, t]}(A_{1}^{r}, \cdots, A_{n}^{r}) \\&\leq& K_{+}(h, r)^{2}K(h, 2)^{r} (G_{[n, t]}(A_{1}, \cdots, A_{n}))^{r}.\end{eqnarray*}$

Proof  If $0<r\leq 1$, then Theorem $C$ implies that

$\begin{equation} K(h, 2)G_{[n, t]}(A_{1}, \cdots, A_{n}) \geq A_{[n, t]}(A_{1}, \cdots, A_{n}) \geq G_{[n, t]}(A_{1}, \cdots, A_{n}). \end{equation}$

(3.2)

By raising all terms of (3.2) to the power $r$, and notice that $t^{r}$ is operator concave for $0<r\leq1$, then we have

$\begin{eqnarray*}K(h, 2)^{r}(G_{[n, t]}(A_{1}, \cdots, A_{n}))^{r} &\geq& (A_{[n, t]}(A_{1}, \cdots, A_{n}))^{r} \\&\geq& A_{[n, t]}(A_{1}^{r}, \cdots, A_{n}^{r}) \\&\geq& G_{[n, t]}(A_{1}^{r}, \cdots, A_{n}^{r}). \end{eqnarray*}$

Next, replace $A_{i}$ by $A_{i}^{r}$ in (3.2) for all $i=0, 1, \cdots, n$, it follows from Theorem 3.1 that

$\begin{eqnarray*}K(h^{r}, 2)G_{[n, t]}(A_{1}^{r}, \cdots, A_{n}^{r}) &\geq& A_{[n, t]}(A_{1}^{r}, \cdots, A_{n}^{r}) \\&\geq& K_{+}(h^{r}, \frac{1}{r})^{-r}A_{[n, t]}(A_{1}, \cdots, A_{n})^{r} \\&\geq& K_{+}(h^{r}, \frac{1}{r})^{-r}(G_{[n, t]}(A_{1}, \cdots, A_{n}))^{r}.\end{eqnarray*}$

Therefore, we have

$K_{+}(h^{r}, \frac{1}{r})^{-r}K(h^{r}, 2)^{-1}(G_{[n, t]}(A_{1}, \cdots, A_{n}))^{r} \leq G_{[n, t]}(A_{1}^{r}, \cdots, A_{n}^{r}).$

If $r\geq1$, then, it follows from (3.2) and Theorem B that

$\begin{equation}K_{+}(h, r)K(h, 2)^{r}(G_{[n, t]}(A_{1}, \cdots, A_{n}))^{r} \geq (A_{[n, t]}(A_{1}, \cdots, A_{n}))^{r}, \end{equation}$

(3.3)

which leads to the following inequalities by (ⅱ) of Theorem 3.1 that

$\begin{equation}(A_{[n, t]}(A_{1}, \cdots, A_{n}))^{r} \geq \frac{1}{K_{+}(h, r)}(A_{[n, t]}(A_{1}^{r}, \cdots, A_{n}^{r})) \geq \frac{1}{K_{+}(h, r)}(G_{[n, t]}(A_{1}^{r}, \cdots, A_{n}^{r})).\end{equation}$

(3.4)

(ⅱ) Replace $A_{i}^{r}$ by $A_{i}$ in (3.3), since $t^{r}$ is operator convex for $1<r\leq2$, we have

$\begin{eqnarray*}K(h^{r}, 2)G_{[n, t]}(A_{1}^{r}, \cdots, A_{n}^{r}) &\geq& A_{[n, t]}(A_{1}^{r}, \cdots, A_{n}^{r})\\&\geq& A_{[n, t]}(A_{1}, \cdots, A_{n})^{r} \\&\geq& \frac{1}{K_{+}(h, r)}G_{[n, t]}(A_{1}, \cdots, A_{n})^{r}.\end{eqnarray*}$

On the other hand, we can get the second inequality from (3.4) and (3.5) immediately.

(ⅲ) If $r>2$, it follows from (3.3), (3.4) and (3.5) that

$\begin{eqnarray*}K(h^{r}, 2)G_{[n, t]}(A_{1}^{r}, \cdots, A_{n}^{r}) &\geq& A_{[n, t]}(A_{1}^{r}, \cdots, A_{n}^{r})\\&\geq& \frac{1}{K_{+}(h, r)}A_{[n, t]}(A_{1}, \cdots, A_{n})^{r} \\&\geq& \frac{1}{K_{+}(h, r)^{2}}G_{[n, t]}(A_{1}, \cdots, A_{n})^{r}.\end{eqnarray*}$

The second inequality follows from (3.4) and (3.5) immediately.