Weighted AM-GM is usually stated as follows:

given the non-negative reals $a_1,a_2,\dots,a_n$ and $\omega_1,\omega_2,\dots,\omega_n\ge 0$ with $\omega_1+\omega_2+\dots+\omega_n=1$ we have:$$\omega_1 a_1+\omega_2 a_2+\dots+\omega_n a_n\ge a_1^{\omega_1} a_2^{\omega_2}\dots a_n^{\omega_n}$$

Now, I was reading Mildorf's introduction to inequalities, he applies weighted AM-GM as follows:$$x_1+\frac{x_2^2}{2}+\frac{x_3^3}{3}+\dots+\frac{x_n^n}{n}\ge (1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n})\cdot\:\sqrt[1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}]{x_1x_2x_3...x_n}$$ How does he apply weighted AM-GM to obtain that?

  • $\begingroup$ Let $H_n:=\sum_{i=1}^n 1/i$. Is your right-hand side meant to be $(H_n^{H_n})(\prod_i x_i)^{1/2}$, or $H_n(\prod_i x^i)^{1/H_n}$? $\endgroup$
    – J.G.
    Dec 7, 2018 at 14:44
  • $\begingroup$ @J.G. The second one, sorry for the unclear LaTeX, it's $H_n\cdot \sqrt[H_n]{x_1x_2x_3...x_n}$ i.e. the $H_n$-th root of the product $\endgroup$ Dec 7, 2018 at 14:52

1 Answer 1


A proof written while discovering the solution:

Define $H_n:=\sum_{i=1}^n\frac{1}{i}$ so the desired result is $\sum_i\dfrac{x_i^i}{iH_n}\ge\prod_ix_i^{1/H_n}$. Choosing weights $\omega_i$, the right-hand side is the weighted GM of the $x_i^{1/(\omega_i H_n)}$. Choose $\omega_i=\frac{1}{iH_n}$, so $\sum_i\omega_i=1$; the weighted GM of the $x_i^{1/\omega_i H_n}$ is then $\sum_i\dfrac{x_i^i}{iH_n}$ as required.

A neater version of the proof that doesn't show the discovery strategy:

With $\omega_i=\frac{1}{iH_n}$ and $x_i^i$ instead of $x_i$ throughout, $\sum_i\dfrac{x_i^i}{iH_n}\ge\prod_ix_i^{1/H_n}$ immediately follows; then multiply by $H_n$.


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