Denote by $\phi$ the golden ratio then show the inequality :
$$\int_{0}^{\phi}f\left(x^{\phi}\right)+f\left(\phi^{\phi}-x\right)dx<\phi$$
Where :
$$f\left(x\right)=\left(\frac{1}{x+1}\right)^{\frac{1}{x}}$$
My attempt :
Using the concavity of $f(x)$ on the domain of the integral we have :
$$f(x)\leq f'\left(\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1+\sqrt{5}}{2}\right)+f\left(\frac{1+\sqrt{5}}{2}\right)$$
Which is currently not sufficient . We can use two derivative instead but by hand it becomes tedious .
Second attempt :
Using logarithm and derivative on the domain of the integral we have :
$$g(x)=\left(\frac{x+1}{x+7}\right)^{\frac{1}{2}}+\frac{1-\sqrt{1+\left(x-1\right)\left(x-2\right)}}{\frac{16}{16\ln\left(2\right)-11}}\geq f(x)$$
Using a computer we have :
$$\int_{0}^{1}\left(f\left(x^{a}\right)+f\left(a^{a}-x\right)\right)dx+\int_{1}^{a}\left(g\left(x^{a}\right)+g\left(a^{a}-x\right)\right)dx<\phi=a$$
Then we have this monster
Last attempt :
Again on the domain of the integral we have :
$$h(x)=\frac{4}{5}\left(\frac{x+1}{x+4}\right)-\frac{1}{5}+e^{-1}+\frac{7\sqrt{x}}{94\cdot2e}-\frac{x^{2}\left(x-1\right)}{256}\geq \left(\frac{1}{x+1}\right)^{\frac{1}{x}}$$
Then with a computer we have :
$$\int_{0}^{\phi}\left(h\left(x^{\phi}\right)+f\left(\phi^{\phi}-x\right)\right)dx<\phi$$
Can we pretend to show it by hand without the use of a computer ? If too hard can we show it analytically with the help of a computer for computing some constant ?
IMHO, the answer is : hopefuly not by hand because this inequality is very tight as you can see on this figure, where we have plotted the curve of $f$ (in green) and the curve of the integrand function (blue curve):
$$y=g(x)=f(x^{\varphi})+f(\varphi^{\varphi}-x)$$
The mean value $m$ of $g$ looks to be very close to $1$ on interval $(0, \varphi)$, giving an integral very close to $m \times \text{interval length}$; i.e., close very close to $1(\varphi - 0)=\varphi$.