# How to prove inequality $a^{\frac{1}{b}}+b^{\frac{1}{a}}\leq \frac {1}{2}$ if $0 <a,b < 1$ and $a+b=1$?

Suppose that $$0 and $$a+b=1$$

Today, I did some investigation of the expression $$a^{\frac{1}{b}}+b^{\frac{1}{a}}$$ and it seems that the maximum is at $$a=b=\frac{1}{2}$$ where the expression is equal to $$\frac {1}{2}$$.

I would like to conjecture that we have $$a^{\frac{1}{b}}+b^{\frac{1}{a}}\leq \frac {1}{2}$$

Is this true?

• @Macavity Where did you find greater value? – Grešnik Jun 24 at 16:15

Consider the concave function $$f(t) = \sqrt[t]{1-t}$$ (shown below.) Using Jensen’s inequality,
$$f(a) + f(b) \leqslant 2f(\tfrac12) = \tfrac12$$
To show $$f$$ is concave, it is enough to show $$g=\log f$$ is concave, as $$t\mapsto e^t$$ is convex and increasing. Perhaps the easiest way for this is to note the Taylor series for $$x \in (0, 1)$$ for $$g= -\sum_{n \geqslant 0} \frac{x^n}{n+1}$$ which implies all coefficients of $$g’’$$ are going to be negative as well $$\implies g’’<0$$.
• Not exactly, but there are some compositions with monotone functions which provide deterministic results. For e.g. if $h$ is convex, decreasing, and $g$ is concave, then $h(g)$ is convex. Helps identify convexity sometimes when derivatives are more difficult, as is the present case. – Macavity Jun 24 at 16:45
• @Macavity I think, to prove by hand that $f$ is a concave function it's not so easy. – Michael Rozenberg Jun 24 at 17:29
• @MichaelRozenberg Have added a note above which explicitly shows $\log f$ Is concave, which is enough to conclude $f$ is. – Macavity Jun 24 at 18:52