[sp]First, I want to prove the inequality $x^x > 1-\sqrt x$, for $0<x<1.$ Taking logs, this is equivalent to $x\ln x > \ln(1-\sqrt x).$
Let $f(x) = x\ln x - \ln(1-\sqrt x).$ Then $f(x)\to0$ as $x\searrow0$. So to prove the inequality it will be enough to show that $f'(x)>0$ for $0<x<1.$ But $f'(x) = 1+\ln x + \dfrac1{2\sqrt x(1-\sqrt x)} > 1+\ln x + \dfrac1{2\sqrt x}$, because $\frac 1{1 - \sqrt x} > 1.$ So it will be enough to show that $g(x)>0$, where $g(x) = 1+\ln x + \dfrac1{2\sqrt x}$.
Differentiating again, $$g'(x) = \frac1x - \frac1{4x^{3/2}},$$ which has a single zero when $x = \frac1{16}$. This turning point is a local minimum (as you can see by differentiating again), and $g\bigl(\frac1{16}\bigr) = 3-4\ln2 \approx 0.2274 > 0$. It follows that $g(x)>0$ throughout the interval $(0,1)$, as required. Working backwards, this means that the inequality $x^x > 1-\sqrt x\;\; (0<x<1)$ is proved.
Now we can look at the inequality $a^b + b^a > 1.$ It is obviously satisfied if either $a$ or $b$ is greater than $1$, so we may assume that they are both less than $1$. Next, notice that when $a,b \in (0,1)$, $a^b$ increases when $a$ increases, and it decreases when $b$ increases.
Suppose that $b < \frac12.$ Then $b^a > a^a > 1-a^{1/ 2}$ (by the above inequality). But $a^{1/ 2} < a^b$, and so $1-a^{1/ 2} > 1-a^b.$ Thus $b^a > 1-a^b$, so that the inequality $a^b+b^a >1$ is proved.
The same argument shows that the inequality holds when $a < \frac12.$ So we only have to look at the case when both $a$ and $b$ lie between $\frac12$ and $1$. But then $a^b+b^a > a+b >1.$ That completes the proof.[/sp]
