Prove or disprove $f(x)+f(f(x))\geq 2$ Let $x>0$ and  then define :
$$\left(x^{-e^{\left(x-1\right)}}\right)=f(x)$$
Claim :
$$f(x)+f(f(x))\geq 2$$
My attempt :
The claim seems not hard on $[1,\infty)$ because we can separate in two the inequality so it seems we have :
$$2-x-f(x)\leq0$$
And :
$$x-f(f(x))\leq0$$
Unfortunately I'm stuck to show it .
Edit Some generalisation :
Let $x>0$ then it seems we have :
$$f\left(x\right)+f\left(f\left(x\right)\right)+f\left(f\left(f\left(x\right)\right)\right)+f\left(f\left(f\left(f\left(x\right)\right)\right)\right)\geq 4$$
Conjecture :
Let $x>0$ then it seems we have :
$$f(x)+f(f(x))+\cdots+f^{(2n)}(x)\geq 2n$$
Where $(2n)$ means the composition not the derivatives and $f^{(1)}(x)=f(x)$.
Question :
How to prove or disprove the claim ?
Thanks you !
 A: Let $y = f(x)$ and $z = f(x) + f(f(x)) = y + y^{-e^{y-1}}$
We want to show $z$ is minimum at $x = 1$
$\frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx} = \left(1 - \left(y^{-e^{y-1}}e^{y-1}(\ln y  + \frac{1}{y})\right)\right)\frac{dy}{dx} = a * \frac{dy}{dx} = 0$
$a = 0$ at $y = 1$ and $y = 1 \implies x = 1$
We can show its second derivative is positive. So $z \ge 2$
A: Since $f(x)>0$ for all $x>0$, it suffices to prove the inequality
$$y+f(y)\geq 2$$
for all $y>0$. In other words, we need to show
$$y^{-e^{y-1}}\geq 2-y.$$
This is clear for $y\geq 2$. Otherwise, taking logarithms, we need to show
$$-e^{y-1}\ln(y)\geq \ln(2-y)$$
for $0<y<2$. Now, substitute $y=1+t$ for $-1<t<1$ to get that we need
$$-e^t\ln(1+t)\geq \ln(1-t)\Leftrightarrow e^{t/2}\ln(1+t)+e^{-t/2}\ln(1-t)\leq 0.$$
Writing
$$e^{t/2}\ln(1+t)=\sum_{k=0}^\infty a_kt^k,$$
we need
$$\sum_{\ell=0}^\infty 2a_{2\ell}t^{2\ell}\leq 0,$$
so it suffices to show that $a_{2\ell}\leq 0$ for all integers $\ell\geq 0$. (The reason the original inequality is so sharp, especially in the neighborhood of $y=1$, is because $a_0=a_2=0$.) Using the Taylor series expansions for $e^x$ and $\ln(1+x)$, we can write
$$-a_{2\ell}=\sum_{k=0}^{2\ell-1}\frac{(-1/2)^k}{k!(2\ell-k)}=\sum_{j=0}^{\ell-1}\left(\frac{1}{2^{2j}(2j)!(2\ell-2j)}-\frac1{2^{2j+1}(2j+1)!(2\ell-2j-1)}\right).$$
So, it will be enough to show
$$\frac{1}{2^{2j}(2j)!(2\ell-2j)}\geq \frac1{2^{2j+1}(2j+1)!(2\ell-2j-1)}$$
for $0\leq j<\ell$. Cancelling like terms, this is equivalent to
$$(2\ell-2j-1)(2j+1)\geq \ell-j,$$
which holds since
$$(2\ell-2j-1)(2j+1)\geq 2\ell-2j-1=(\ell-j)+(\ell-j-1)\geq \ell-j.$$
This finishes the proof.
Note that your generalization to sums of repeated applications of $f$ follows from applying the inequality $x+f(x)\geq 2$ at various values of $x$, while your intermediate conjecture $x\leq f(f(x))$ is false for $x<1$.
