TITLE: If $a+b=2$ so $a^{a^{\frac{b}{2}}}b^{b^{\frac{a}{2}}}\geq1$
QUESTION [6 upvotes]: Let $a$ and $b$ be positive numbers such that $a+b=2$. Prove that:
  $$a^{a^{\frac{b}{2}}}b^{b^{\frac{a}{2}}}\geq1$$

We can prove an easier inequality $a^{a^{b}}b^{b^{a}}\geq1$ with the same conditions by the following reasoning.
Let $a\geq1\geq b>0$. Hence, $a^{b-1}\geq b^{a-1}$, which gives $a^b\geq\frac{a}{b}\cdot b^a$ and
$$a^{a^{b}}b^{b^{a}}\geq a^{\frac{a}{b}\cdot b^a}b^{b^{a}}.$$
Thus, it remains to prove that
$$a^{\frac{a}{b}\cdot b^a}b^{b^{a}}\geq1$$ or
$$a^ab^b\geq1,$$ which follows from
$$a^ab^b\geq\left(\frac{a+b}{2}\right)^{a+b},$$
which is Jensen for $f(x)=x\ln{x}$.

REPLY [2 votes]: Lemma 1. If $t \in [0,1]$ and $x \in [0,1)$, then
$$(1+x)^t \geq 1 + t x + \frac{1}{2} \left(t^2-t\right) x^2.$$
If $t \in [0,1]$ and $x \in (-1,0]$, then
$$(1+x)^t \leq 1 + t x + \frac{1}{2} \left(t^2-t\right) x^2.$$
Proof. Let $t \in [0,1]$. We define the function $f_t\, \colon (-1,1) \rightarrow \mathbb{R}$ as given by
$$f_t(x) = (1+x)^t - \left(1 + t x + \frac{1}{2} \left(t^2-t\right) x^2\right).$$
For all $x \in (-1,1)$, we have $$f_t'(x) = t \left((1+x)^{t-1}-\left(1+(t-1)x\right)\right).$$
By Bernoulli's inequality, $f_t' \geq 0$, thus $f_t$ is increasing. Since $f_t(0) = 0$, we have $f_t(x) \geq 0$ for all $x \in [0,1)$ and $f_t(x) \leq 0$ for all $x \in (-1,0]$.
$$\tag*{$\Box$}$$
We will now prove the given inequality in an equivalent, symmetric form.
Claim. If $x \in (-1,1)$, then $$(1+x)^{(1-x)/2} \log (1+x)+(1-x)^{(1+x)/2} \log (1-x) \geq 0.$$
Proof. By symmetry, it is sufficient to show that the inequality holds on $[0,1)$. Let $x \in [0,1)$. 
By Lemma 1, we have
\begin{align*}
(1+x)^{(1-x)/2} &\geq 1 + \frac{1}{2} (1-x) x + \frac{1}{2} \left(\frac{1}{4} (1-x)^2-\frac{1-x}{2}\right) x^2
\\&= \frac{1}{8} \left(x^4-5 x^2+4 x+8\right),
\\ \text{and} \qquad\qquad&
\\ (1-x)^{(1+x)/2} &\leq  \frac{1}{8} \left(x^4-5 x^2-4 x+8\right).
\end{align*}
Therefore it is sufficient to show that $$\left(x^4-5 x^2+4 x+8\right) \log (1+x)+\left(x^4-5 x^2-4 x+8\right) \log (1-x) \geq 0.$$
We have 
\begin{align*}
&\left(x^4-5 x^2-4 x+8\right) \log (1-x) 
\\[7pt] &= \left(x^4-5 x^2-4 x+8\right) \sum_{k=1}^{\infty} \left(-\frac{x^k}{k}\right)
\\[7pt] &= 
- \sum_{k=1}^{\infty} \frac{x^{k+4}}{k}
+ \sum_{k=1}^{\infty} \frac{5x^{k+2}}{k}
+ \sum_{k=1}^{\infty} \frac{4x^{k+1}}{k}
- \sum_{k=1}^{\infty} \frac{8x^k}{k}
\\[7pt] &= 
- \sum_{k=5}^{\infty} \frac{x^{k}}{k-4}
+ \sum_{k=3}^{\infty} \frac{5x^{k}}{k-2}
+ \sum_{k=2}^{\infty} \frac{4x^{k}}{k-1}
- \sum_{k=1}^{\infty} \frac{8x^k}{k}
\\[7pt] &= -8 x +\frac{13 x^3}{3} + \frac{11 x^4}{6} + 
 \sum_{k=5}^{\infty} \left(-\frac{1}{k-4}+\frac{5}{k-2}+\frac{4}{k-1}-\frac{8}{k}\right) x^k
 \\[7pt] &= -8 x +\frac{13 x^3}{3} + \frac{11 x^4}{6} + 
 \sum_{k=5}^{\infty} \frac{2 \left(5 k^2-31 k+32\right)  x^k}{(k-4) (k-2) (k-1) k}.
\end{align*}
Thus
\begin{align*}
&\left(x^4-5 x^2+4 x+8\right) \log (1+x) + \left(x^4-5 x^2-4 x+8\right) \log (1-x)
\\[7pt] &= \frac{22 x^4}{6} + 
 \sum_{k=3}^{\infty} \frac{4 \left(5\cdot (2k)^2-31 \cdot (2k)+32\right)  x^{2k}}{(2k-4) (2k-2) (2k-1) 2k}
 \\[7pt] &= \frac{22 x^4}{6} + 
 \sum_{k=3}^{\infty} \frac{4 \left(10 k^2-31 k+16\right) x^{2k}}{(2k-4) (2k-2) (2k-1) k}.
\end{align*}
Since $10 k^2-31 k+16 \geq 0$ for all $k \geq 3$, we are done.
$$\tag*{$\Box$}$$