Inequality of convex plus concave function Given $\displaystyle f(x) = \sqrt[x]{1+x}$, prove:
$$f(x)+f\left(\frac{1}{x}\right)<4 \quad \forall x>0$$
My reduction so far: it is sufficient to show the inequality holds for $\displaystyle x \in (0,1]$
Jensen's Inequality can't provide an estimate since the LHS is a sum of a convex and concave function.
Do any of the "standard olympiad tricks" work for pulling out this estimate? Or is there no other means besides calculus?
 A: Since function $g(x)=\frac{\ln(1+x)}{x}$ is decreasing when $x\geq 0$, for $\frac12\leq x\leq 1$ we have
$$f(x)+f(1/x)=e^{g(x)}+e^{g(1/x)}\leq e^{g(\frac12)}+e^{g(1)}=e^{2\ln\frac32}+e^{\ln 2}=\frac92<4.$$
Also, in the case $\frac{1}{10}\leq x<\frac12$ we have
$$f(x)+f(1/x)=e^{g(x)}+e^{g(1/x)}\leq e^{g(\frac{1}{10})}+e^{g(2)}=e^{10\ln\frac{11}{10}}+e^{\frac12\ln3}=\left(\frac{11}{10}\right)^{10}+\sqrt3<4.$$
Finally, for $0<x<\frac{1}{10}$ we have
$$f(x)+f(1/x)=e^{g(x)}+e^{g(1/x)}\leq e^{g(0)}+e^{g(10)}=e+e^{\frac{\ln 11}{10}}=e+\sqrt[10]{11}<4.$$
A: $$f(x)+f\left(\frac{1}{x}\right)\le4$$
is actually:
$$ (1+x)^{\frac1x} + (1+\frac1x)^x\le 4$$
It can be written in the following form:
$$ (1+a)^b + (1+b)^a\le 4,\quad ab=1$$
$$ (1+\frac1b)^b + (1+\frac1a)^a\le 4,\quad ab=1$$
Function $f(x)=(1+\frac1x)^x$ is concave so we can apply Jensen:
$$f(a)+f(b)\le2f(\frac{a+b}{2})$$
$$ (1+\frac1b)^b + (1+\frac1a)^a\le 2\left(1+\frac1{\frac{a+b}{2}}\right)^{\frac{a+b}2}$$
...or by AM-GM:
$$ (1+\frac1b)^b + (1+\frac1a)^a\le 2\left(1+\frac1{\sqrt{ab}}\right)^{\frac{a+b}2}=2\cdot2^{\frac{a+b}2}$$
It's easy to prove that expression $\frac{a+b}2$ with constraint $ab=1$ reaches minimum value of 1 for $a=b=1$. Therefore:
$$LHS\le4$$
...with equality only for $a=b=x=1$.
