# 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?

• Note that $f(1)+f(1/1)=4$. – Robert Z May 31 '18 at 11:30
• You don't have to mix convex and concave functions. Replace $x$ with $1/y$ and introduce additional constraint $xy=1$. Now you have only one, conave function to deal with and you can happily apply Jensen. – Oldboy May 31 '18 at 13:01
• A nice proof is given in math.stackexchange.com/questions/2899031/… – River Li Jul 24 at 13:08

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 $$f(x)+f(1/x)=e^{g(x)}+e^{g(1/x)}\leq e^{g(0)}+e^{g(10)}=e+e^{\frac{\ln 11}{10}}=e+\sqrt[10]{11}<4.$$

• In the first line $e^{2\ln\frac32}+e^{\ln 2}=\frac{17}{4}>4$ Am I right? – Robert Z May 31 '18 at 11:28
• Yes, you're right, it's a flaw. – CuriousGuest May 31 '18 at 18:06

$$f(x)+f\left(\frac{1}{x}\right)\le4$$

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 $a=b=x=1$.

• Yes, it is true that given the contraint that $ab = 1$, the minimum value of $\frac{a + b}{2}$ is equal to $1$. In fact it's just a consequence of the AM-M inequliaty, which you had already cited. The problem, however, is that this implies that $2 \cdot 2^{\frac{a + b}{2}}$ is always greater than or equal to $4$, so you've only shown that $f(x) + f\left(\frac{1}{x}\right)$ is bounded above by something that is larger than $4$, but you haven't shown that it is less than $4$. It could lie somewhere between $4$ and $2 \cdot 2^{\frac{a + b}{2}}$. So you haven't actually proven the inequality. – Dylan May 31 '18 at 23:52
• @Dylan: You're right. This has to be re-worked. I'm asking the submitter to unaccept this answer. Also upvoting your comment. – Oldboy Jun 1 '18 at 6:03
• @Oldboy I have unaccepted the answer – Jack Tiger Lam Jun 3 '18 at 3:02