Prove the inequality $({1+\frac{a}b})^n$ + $(1+\frac{b}a)^n$ $\geq$ $2^{n+1}$

Solution 1:

Using the Arithmetic Mean Geometric Mean Inequality, we have $$\frac{(1+a/b)^n+(1+b/a)^n}{2}\ge \sqrt{((1+a/b)(1+b/a))^n}.$$ But $(1+a/b)(1+b/a)=2+a/b+b/a\ge 4$, since $a/b+b/a\ge 2$, again by AM/GM. The result follows.

Remark: Mentioning AM/GM is perhaps overly fancy. We have used twice the result that for non-negative $x$ and $y$, we have $\frac{x+y}{2}\ge \sqrt{xy}$. This follows from the fact that $x+y-2\sqrt{xy}=(\sqrt{x}-\sqrt{y})^2$.

Solution 2:

$a,b$ are positive, so $$\dfrac{a^k}{b^k}+\dfrac{b^k}{a^k}\ge2\Longleftrightarrow a^{2k}+b^{2k}\ge2a^kb^k\Longleftrightarrow\big(a^k-b^k\big)^2\ge0,$$ and using the binomial theorem (three times total) we get $$\left(1+\frac ab\right)^n+\left(1+\frac ba\right)^n=\sum_{k=0}^n\binom nk\left(\dfrac{a^k}{b^k}+\dfrac{b^k}{a^k}\right)\ge2\sum_{k=0}^n\binom nk=2\,(1+1)^n=2^{n+1}.$$

Solution 3:

Consider $f(x)= (1+x)^n + \left(1+\dfrac1x\right)^n$ for $x>0$.

Then $f'(x)=n(1+x)^{n-1}\left(1-\dfrac1{x^{n+1}}\right)$. So $f'(x)=0$ for $x>0$ iff $x=1$.

Since $f(x)\to +\infty $ when $x\to+\infty$, $\quad f$ has a minimum at $x=1$.

The minimum is $f(1)=2\cdot2^n=2^{n+1}$.