Direct Proofs Example
The arithmetic/geometric mean inequality
Definition: The arithmetic mean of two real numbers \(a\) and \(b\) is \((a+b)/2\).
Definition: The geometric mean of two positive real numbers \(a\) and \(b\) is \(\sqrt(ab)\).
Proposition: If \(a\) and \(b\) are positive real numbers, then the geometric mean of \(a\) and \(b\) is less than or equal to their arithmetic mean.
Proposition: If \(a\) and \(b\) are positive real numbers, then the geometric mean of \(a\) and \(b\) is less than or equal to their arithmetic mean.
Proposition: If \(a\) and \(b\) are positive real numbers, then: \[ \sqrt(ab)\le \frac{(a+b)}{2} \]
Problem Solving Phase
Isolating the needed lemma
Lemma: If \(a\) and \(b\) are positive, and \(a\le b\), then \(\sqrt{a}\le\sqrt{b}\) where \(\sqrt{x}\) denotes the positive square root of \(x\).
The proof
Proposition: If \(a\) and \(b\) are positive real numbers, then: \[ \sqrt(ab)\le \frac{(a+b)}{2} \]
Proof:
We know that \((a-b)^2\ge 0\). Therefore \(a^2+b^2-2ab\ge 0\) and so \(a^2+b^2\ge 2ab\). Add \(2ab\) to both sides to obtain \(a^2+2ab+b^2\ge 4ab\) so \((a+b)^2\ge 4ab\). Both sides of this inequality are positive, since the left side is a square the right side is a product of positive numbers. Now apply the lemma to take the square root of both sides to obtain \[ (a+b)\ge 2\sqrt{ab}. \] Dividing both sides by \(2\) yields the desired result.