We use that $x+\frac{1}{x} \geq 2$ for all $x \in \mathbb{R}_{>0}$. It holds that

$$\begin{aligned} \left(1+\frac{a}{b}\right)^{m}+\left(1+\frac{b}{a}\right)^{m} & =\sum_{i=0}^{m}\binom{m}{i}\left(\frac{a}{b}\right)^{i}+\sum_{i=0}^{m}\binom{m}{i}\left(\frac{b}{a}\right)^{i} \\ & =\sum_{i=0}^{m}\binom{m}{i}\left(\left(\frac{a}{b}\right)^{i}+\left(\frac{b}{a}\right)^{i}\right) \\ & =\sum_{i=0}^{m}\binom{m}{i}\left(\frac{a^{i}}{b^{i}}+\frac{b^{i}}{a^{i}}\right) \\ & \geq \sum_{i=0}^{m}\binom{m}{i} \cdot 2 \\ & =2^{m+1} . \end{aligned}$$

## Alternative solution.

Apply the inequality of the arithmetic and geometric means to 1 and $\frac{a}{b}$:

$$1+\frac{a}{b} \geq 2 \sqrt{\frac{a}{b}}$$

so

$$\left(1+\frac{a}{b}\right)^{m} \geq\left(2 \sqrt{\frac{a}{b}}\right)^{m}$$

Similarly, of course,

$$\left(1+\frac{b}{a}\right)^{m} \geq\left(2 \sqrt{\frac{b}{a}}\right)^{m}$$

Now we apply the inequality of the arithmetic and geometric means again:

$$\left(2 \sqrt{\frac{a}{b}}\right)^{m}+\left(2 \sqrt{\frac{b}{a}}\right)^{m} \geq 2 \sqrt{\left(2 \sqrt{\frac{a}{b}}\right)^{m} \cdot\left(2 \sqrt{\frac{b}{a}}\right)^{m}}=2^{m+1}$$

This proves the desired result.