A weird inequality

One needs to maximise the first part and minimise the second part in order to get maximum value of the whole expression.

For the second part:

$\displaystyle \bigg( \frac{ e^x + e^y + e^z} {3} \bigg)_{min}$

Applying $\displaystyle AM-GM$ and also the condition for equality $\displaystyle x = y = z = 2$ :

$\displaystyle = \sqrt[3]{ e^2 \cdot e^2 \cdot e^2}$

$\displaystyle = e^2$

For the first part:

$\displaystyle \Bigg( \frac{2}{3} \Big( e^{\frac{x}{2}} \cdot e^{\frac{y}{2}} + e^{\frac{y}{2}} \cdot e^{\frac{z}{2}} + e^{\frac{z}{2}} \cdot e^{\frac{x}{2}} \Big) \Bigg)_{max}$

Applying Cauchy-Shwartz inequality and also applying the condition for equality $\displaystyle x = y = z = 2$ :

$\displaystyle = \frac{2}{3} \bigg( \sqrt{ \big( (e^{\frac{x}{2}})^{2} + (e^{\frac{y}{2}})^{2} + (e^{\frac{z}{2}})^{2} \big) \cdot \big( (e^{\frac{x}{2}})^{2} + (e^{\frac{y}{2}})^{2} + (e^{\frac{z}{2}})^{2} \big) } \bigg)$

$\displaystyle = \frac{2}{3} \sqrt{ (3e^{2}) \cdot (3e^{2}) }$

$\displaystyle = \frac{2}{3} \times (3 e^{2})$

$\displaystyle = 2 e^{2}$

Hence, the maximum value of the given expression boils down to:

$\displaystyle M = (2 e^{2} ) - (e^{2})$

$\displaystyle M = e^{2} = 7.3891$

$\displaystyle \Rightarrow \lfloor 1000M \rfloor = \boxed{7389}$

Let $\begin{aligned} X & = \frac 23 \left(\color{#3D99F6}{e^\frac {x+y}2 + e^\frac {y+z}2 + e^\frac {z+x}2}\right) - \frac {\color{#D61F06}{e^x+e^y+e^z}}3 \end{aligned}$

It is shown later that when $x=y=z=2$ , $X$ maximizes when $\color{#3D99F6}{Y = e^\frac {x+y}2 + e^\frac {y+z}2 + e^\frac {z+x}2}$ is maximum and $\color{#D61F06}{Z = e^x+e^y+e^z}$ is minimum.

Using RMS-AM inequality on $Y$

$\begin{aligned} \frac {\color{#3D99F6}{e^\frac {x+y}2 + e^\frac {y+z}2 + e^\frac {z+x}2}}3 & \le \sqrt{\frac {e^{x+y} + e^{y+z} + e^{z+x}}3} & \small \color{#3D99F6}{\text{At equality }x=y=z=2} \\ & \le e^2 \end{aligned}$

Applying AM-GM inequality on $Z$

$\begin{aligned} \frac {\color{#D61F06}{e^x+e^y+e^z}}3 & \ge \sqrt[3]{e^{x+y+z}} = \sqrt[3]{e^6} = e^2 \end{aligned}$

Again equality occurs when $x=y=z=2$ .

Therefore, $X_{max} = 2e^2 - e^2 = e^2 \approx 7.389$ $\implies \lfloor 1000M\rfloor = \lfloor 1000\times 7.389\rfloor = \boxed{7389}$

consider the function $f(a)=e^{a}$ now use this .