Prove $(x+r_1) \cdots (x+r_n) \geq (x+(r_1 \cdots r_n)^{1/n})^{n}$. I can show that for $x > 0$ and $r_{i} > 0$ we have
$$
\left(\, x + r_{1}\,\right)\ldots\left(\, x + r_{n}\,\right)\
\geq\
\left[\, x + \left(\, r_{1}\ldots r_{n}\,\right)^{1/n}\,\right]^{n}.$$
However, I can't do this using straight up induction, strong or weak. Can someone do this? 
You can show the inequality holds for $n$ if it holds for $n+1$ using $r_{n+1} = (r_1 \cdots r_n)^{1/n}$, so going down isn't a problem, but going up has eluded me. Additionally, you can show that if it holds for $n=a$ and $n=b$ then it holds for $n=ab$. Using powers of two and the above is sort of a way to prove the above by induction but it certainly isn't "normal".
You can also prove the above using Lagrange multipliers (which isn't surprising). The last proof I know is where you compare coefficients of $x^k$ and use the AM-GM inequality.

Here's a proof by induction: It holds trivially for $n=1$. For $n\ge2$ we have
\begin{align*}(x+r_1)&\ldots(x+r_{n-1})(x+r_n)\ge(x+(r_1\ldots r_{n-1})^{1/(n-1)})^{n-1}(x+r_n)\\&=\Bigl(\bigl((x^{(n-1)/n})^{n/(n-1)}+((r_1\ldots r_{n-1})^{1/n})^{n/(n-1)}\bigr)^{(n-1)/n}\bigr((x^{1/n})^n+(r_n^{1/n})^n\bigr)^{1/n}\Bigr)^n\\&\ge\bigl(x+(r_1\ldots r_n)^{1/n}\bigr)^n.\end{align*}
The last step is Hölder's inequality for $p=\frac{n-1}n$ and $q=\frac1n$.
