Prove or disprove $f(x)+f(f(x))\geq 2$

Question

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 !

Answer 1

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 $x<1$.

Answer 2

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 $z \ge 2$