Relation Between A.M. and G.M.

Let $a$ and $b$ be two positive real numbers, with arithmetic mean $A=\dfrac{a+b}{2}$ and geometric mean $G=\sqrt{ab}$ (Sections 3 and 5). A fundamental and constantly-used fact relates the two:

with equality holding if and only if $a=b$.

Proof. Since $\sqrt a$ and $\sqrt b$ are well-defined real numbers (as $a,b>0$), consider the real number $\sqrt a - \sqrt b$. The square of any real number is never negative, so

Expanding the square,

which is exactly $A\ge G$. Moreover, $(\sqrt a-\sqrt b)^2=0$ precisely when $\sqrt a=\sqrt b$, i.e. when $a=b$ — so the inequality is an equality exactly in that one case, and strict ($A>G$) whenever $a\ne b$. $\blacksquare$

Recovering $a$ and $b$ from $A$ and $G$. Since $A=\frac{a+b}{2}$ and $G^2=ab$, the two original numbers satisfy

Whenever the sum and the product of two numbers are known, those numbers are precisely the two roots of the quadratic equation $x^2 - (\text{sum})x + (\text{product}) = 0$ (this is immediate from expanding $(x-a)(x-b)=x^2-(a+b)x+ab$). So $a$ and $b$ are the roots of

This gives a direct method for finding two positive numbers from their given A.M. and G.M. — solve this quadratic by the quadratic formula, $x = A \pm \sqrt{A^2-G^2}$. …