A.M., G.M., H.M. inequality 算數-幾何-調和平均值不等式(Part 1)

\[there\ are\ n\ numbers,\ x_1,\ x_2,\ ....\ ,\ x_n\ ,\ \ \forall i,\ 1\le i\le n,\ x_i\ge0\]

Arithmetic mean (A.M.)算術平均值

\[A.M.\ =\frac{x_1+x_2+...\ +x_n}{n}\]

Geometric mean (G.M.) 幾何平均值

\[G.M.\ =\left(x_1x_2...x_n\right)^{\frac{1}{n}}\]

Harmonic mean (H.M.) 調和平均值

\[H.M.=\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}}\]

\[A.M\ge G.M.\ge H.M\]

Part 1

\[證明\ A.M\ge G.M.\]

\[Let\ the\ statement\ P\left(n\right)\ is\ that\ A.M\ge G.M.\ for\ all\ n\ge2\ where\ n\ is\ the\ number\ ofpositive\ integer\ \]

\[When\ n=2\]

\[\left(\sqrt{a_1}-\sqrt{a_2}\right)^2\ge0\]

\[a_1-2\sqrt{a_1a_2}+a_2\ge0\]

\[\frac{a_1+a_2}{2}\ge\sqrt{a_1a_2}\]

\[P\left(2\right)\ holds\]

\[Assume\ P\left(2^k\right)holds\]

\[i.e.\ \frac{a_1+a_2+...\ +a_{2^k}}{2^k}\ge\left(a_1a_2...a_{2^k}\right)^{\frac{1}{2^k}}\]

\[When\ n=2^{k+1}\]

\[\frac{a_1+a_2+...\ +a_{2^{k+1}}}{2^{k+1}}=\frac{\frac{a_1+a_2+...\ +a_{2^k}}{2^k}+\frac{a_{2^k+1}+a_{2^k+2}+...\ +a_{2^{k+1}}}{2^k}}{2}\]

\[\ge\frac{\left(a_1a_2...a_{2^k}\right)^{\frac{1}{2^k}}+\left(a_{2^k+1}a_{2^k+2}...a_{2^{k+1}}\right)^{\frac{1}{2^k}}}{2}\ge\left(a_1a_2...a_{2^{k+1}}\right)^{\frac{1}{2^{k+1}}}\]

\[\frac{a_1+a_2+...\ +a_{2^{k+1}}}{2^{k+1}}\ge\left(a_1a_2...a_{2^{k+1}}\right)^{\frac{1}{2^{k+1}}}\]

\[P\left(2^k\right)\Rightarrow P\left(2^{k+1}\right)\]

\[Assume\ P\left(k\right)\ holds\]

\[i.e.\ \frac{a_1+a_2+...\ +a_k}{k}\ge\left(a_1a_2...a_k\right)^{\frac{1}{k}}\]

\[When\ n=k-1\]

\[\frac{a_1+a_2+...\ +a_k}{k}\ge\left(a_1a_2...a_k\right)^{\frac{1}{k}}\]

\[Let\ a_k=\frac{a_1+a_2+...\ +a_{k-1}}{k-1}\]

\[\frac{\left(k-1\right)a_k+a_k}{k}\ge\left(a_1a_2...a_k\right)^{\frac{1}{k}}\]

\[a_k\ge\left(a_1a_2...a_k\right)^{\frac{1}{k}}\]

\[a_k^{\frac{k-1}{k}}\ge\left(a_1a_2...a_{k-1}\right)^{\frac{1}{k}}\]

\[\frac{a_1+a_2+...\ +a_{k-1}}{k-1}\ge\left(a_1a_2...a_{k-1}\right)^{\frac{1}{k-1}}\]

\[P\left(k\right)\Rightarrow P\left(k-1\right)\]

\[By\ mathematical\ induction,\ A.M\ge G.M.\ for\ all\ n\ge2\]