If $a+b=2$ so $a^{a^{\frac{b}{2}}}b^{b^{\frac{a}{2}}}\geq1$

Question

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 $f(x)=x\ln{x}$.

Answer 1

If you need to prove that $$A=a^ab^b\geq1\qquad \text{with}\qquad a+b=2\qquad \text{and}\qquad a\geq1\geq b>0$$ let $a=1+x$ and $b=1-x$.

This makes $$A(x)=(1+x)^{1+x}(1-x)^{1-x} $$ $$\log(A(x))=(1+x)\log(1+x)+(1-x)\log(1-x)$$ Differentiating $$\frac{A'(x)}{A(x)}=\log(1+x)-\log(1-x)$$ which is positive if $x>0$.

Answer 2

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. $$\tag*{$\Box$}$$