ProofOfArithmeticGeometricHarmonicMeansInequality2 2004-11-17 12:42:58 2006-06-06 01:59:03 Proof arithmetic-geometric-harmonic means inequality 0 \usepackage{pstricks} A short geometrical proof can be given for the case $n=2$ of the arithmetic-geometric means inequality. Let $a$ and $b$ be two non negative numbers. Draw the line $AB$ such that $AP$ has length $a$, and $PB$ has length $b$, as in the following picture, and draw a semicircle with diameter $AB$. Let $O$ be the center of the circle. \begin{center} \begin{pspicture}(4,4)(-5,-1) \psset{unit=0.7cm} \psset{linewidth=0.5mm} \psarc(0,0){4}{0}{180} \qline(-4,0)(4,0) \qline(-1.5,0)(-1.5,3.708) \uput[ul](-1.5,3.708){$Q$} \uput[d](0,0){$O$} \uput[d](-4,0){$A$} \uput[d](4,0){$B$} \uput[d](-1.5,0){$P$} \uput[u](-2.7,0){$a$} \uput[u](2.7,0){$b$} \end{pspicture} \end{center} Now raise perpendiculars $PQ$ and $OT$ to $AB$. Notice that $OT$ is a radius, and so \[OT=\frac{AB}{2}=\frac{a+b}{2}\] Also notice that $PQ\leq OT$ for any point $P$, and equality is obtained only when $P=O$, that is, when $a=b$. \begin{center} \begin{pspicture}(4,4)(-5,-1) \psset{unit=0.7cm} \psset{linewidth=0.5mm} \psarc(0,0){4}{0}{180} \qline(-4,0)(4,0) \qline(-1.5,0)(-1.5,3.708) \uput[ul](-1.5,3.708){$Q$} \uput[d](0,0){$O$} \uput[d](-4,0){$A$} \uput[d](4,0){$B$} \uput[d](-1.5,0){$P$} \uput[u](0,4){$T$} \qline(-4,0)(-1.5,3.708) \qline(4,0)(-1.5,3.708) \qline(0,0)(0,4) \end{pspicture} \end{center} Notice also that $PQ$ is a height over the hypotenuse on right triangle $\triangle AQB$. We have then triangle similarities $\triangle AQB \sim \triangle APQ\sim \triangle QPB$, and thus \[ \frac{AP}{PQ} = \frac{PQ}{PB} \] which implies $PQ=\sqrt{AP\cdot PB}=\sqrt{ab}$. Since $PQ\leq OT$, we conclude \[ \sqrt{ab}\leq\frac{a+b}{2}. \] \bigskip This special case can also be proved using rearrangement inequality. Let $a,b$ non negative numbers, and assume $a\leq b$. Let $x_1=\sqrt{a},x_2=\sqrt{b}$, and then $x_1\leq x_2$. Now suppose $y_1$ and $y_2$ are such that one of them is $x_1$ and the other is $x_2$. Rearrangement inequality states that $x_1y_1+x_2y_2$ is maximum when $y_1\leq y_2$ and $x_1\leq x_2$. So, we have \[x_1x_2+x_2x_1 \leq x_1^2 + x_2^2\] and substituting back $a,b$ gives \[2 \sqrt{ab} \leq (\sqrt{a})^2 + (\sqrt{b})^2 = a+b\] where it follows the desired result. \bigskip One more proof can be given as follows. Let $x=\sqrt{a}, y =\sqrt{b}$. Then $(x-y)^2 \geq 0$, and equality holds only when $x=y$. Then, $x^2-2xy+ y^2 \geq 0$ becomes \[x^2 + y^2 \geq 2 xy\] and substituting back $a,b$ gives the desired result as in the previous proof.