All is in the title so we want to show that :
Let $a,b\geq e$ then we have
$$\operatorname{W}(a)+\operatorname{W}(b)\leq 2\operatorname{W}\Big(\frac{a+b}{2}\Big)$$
The trick is to put :
$a=x\ln(x)$
$b=y\ln(y)$
$\frac{a+b}{2}=c\ln(c)$
Recalling that $z\geq e^{-1}$:
$$\operatorname{W}(z\ln(z))=\ln(z)$$
We get :
$$\ln(x)+\ln(y)\leq 2\ln(c)$$
But using Jensen's inequality we have $x,y>0$:
$$\ln(x)+\ln(y)\leq 2\ln\Big(\frac{x+y}{2}\Big)$$
So :
$$2\ln\Big(\frac{x+y}{2}\Big)\leq 2\ln(c)$$
Or :
$$\frac{x+y}{2}\leq c$$
Or :
$$\Big(\frac{x+y}{2}\Big)\ln\Big(\frac{x+y}{2}\Big)\leq c\ln(c)$$
Or :
$$\Big(\frac{x+y}{2}\Big)\ln\Big(\frac{x+y}{2}\Big)\leq \frac{x\ln(x)+y\ln(y)}{2}$$
Wich is true again by Jensen's inequality and as $x,y,c\geq e$ and convexity of $f(u)=u\ln(u)$.
So it's mid-point concave and continuous (as the inverse function of the Lambert's function is continuous ) ,so it's concave .
Question
Using a similar argument how to show it for $e\geq a\geq b\geq 0$ ?
Thanks in advance .
Best Answer
Changing notations, let $a=x+\epsilon$ and $b=x-\epsilon$ and use series expanion around $\epsilon=0$. This gives $$W(x+\epsilon)+W(x-\epsilon)=2 W(x)-\Big[\frac{W(x)}{x (W(x)+1)}\Big]^2\frac{W(x)+2}{ W(x)+1}\epsilon ^2 +O\left(\epsilon ^4\right)$$ which seems to be sufficient.
Edit
May be could be used a very recent identity $$W(a)+W(b)=W\left(a b \left(\frac{1}{W(a)}+\frac{1}{W(b)}\right)\right)$$ $$2W\Big(\frac{a+b}{2}\Big)=W\left(\frac{(a+b)^2}{2 W\left(\frac{a+b}{2}\right)}\right)$$ and since $W(t)$ is an incresing function, show the inequality based on the arguments.
Another idea could be to use the infinite series $$W(t)=\sum_{n=1}^\infty (-1)^n\frac{ n^{n-2} }{(n-1)!}t^n$$