# Application of weighted AM-GM

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?

• 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 $H_n(\prod_i x^i)^{1/H_n}$?
– J.G.
Dec 7 '18 at 14:44
• @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 Dec 7 '18 at 14:52

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.
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 $$H_n$$.