Relationship between AM and GM

Recall the two means for two positive real numbers $a$ and $b$:

where: $a, b$ are positive real numbers, $\text{AM}$ is their Arithmetic Mean (the ordinary average), and $\text{GM}$ is their Geometric Mean (the side of a square whose area equals the rectangle $a\times b$). The AM–GM relation states that these two means always sit in a fixed order:

where: equality holds if and only if $a=b$ — the two means coincide precisely when the two numbers are equal, and separate the moment $a$ and $b$ differ. A useful picture to hold in mind: if $a \le b$, the Geometric Mean always lands between the two numbers and below the Arithmetic Mean, giving the chain $a \le \text{GM} \le \text{AM} \le b$. The figure below places this on a number line for $a=9$ and $b=16$, the same pair used in the worked example ahead.

This inequality needs nothing beyond $a,b>0$ (so that $\sqrt a$ and $\sqrt b$ are real numbers) — it is a direct consequence of a perfect square being non-negative, proved next. It is worth noticing where this result sits in the chapter's logic. Arithmetic Mean was first met as the single term inserted between two numbers to keep an Arithmetic Progression running smoothly; Geometric Mean played the same role for a Geometric Progression. The AM–GM relation is the natural next question once both means are on the table: given the same two starting numbers, which of the two inserted means is bigger? The algebra says the answer is always the same — Arithmetic Mean never loses — and that fixed ordering is precisely what makes the relation useful as a tool rather than just a curiosity: it lets you replace an unknown product $\sqrt{ab}$ by the more manageable sum $\tfrac{a+b}{2}$ in an inequality, or vice versa, whenever a "greater than or equal to" is all that is needed.

Problem. Verify the AM–GM relation for $a=9$ and $b=16$, state whether equality holds, and use the same relation to prove that for every positive real number $x$, $x+\dfrac1x \ge 2$.

Step 1 — compute AM and GM for the given pair.

Step 2 — compare. Since $12.5 \ge 12$, the relation $\text{AM} \ge \text{GM}$ holds for this pair. Equality does not hold, because $9 \ne 16$.

Step 3 — apply the relation to a new pair. Take the two positive numbers to be $x$ and $\dfrac1x$ (valid for any $x>0$, since then $\tfrac1x>0$ too). Their means are

Step 4 — apply $\text{AM} \ge \text{GM}$ and clear the fraction.

The final answer: $\text{AM}=12.5$, $\text{GM}=12$, equality does not hold since $a\ne b$; and $x+\tfrac1x \ge 2$ for every positive real $x$, with equality exactly when $x=\tfrac1x$, i.e. $x=1$.

Competitive stretch

The same trick generalises: for any two positive quantities $p$ and $q$ whose product is a fixed constant $k$ (i.e. $pq=k$), AM–GM gives $\tfrac{p+q}{2}\ge\sqrt{k}$, so their sum $p+q$ is minimised — and equals $2\sqrt k$ — exactly when $p=q$. This "fix the product, minimise the sum" idea (or its mirror, "fix the sum, maximise the product") is the standard entry point for optimisation-style inequality questions built on AM–GM.

• Applying AM ≥ GM to non-positive numbers. The relation is proved only for $a,b>0$; if either number is negative or zero, $\sqrt{ab}$ may not be a real number, so the inequality does not apply as stated.
• Forgetting the equality condition. Many questions specifically ask when equality holds — the answer is always "when $a=b$," and it is often the key step in optimisation problems like the $x+\tfrac1x$ example.
• Mixing up which mean is larger. A quick sanity check with $a=1,b=9$ gives $\text{AM}=5 \ge \text{GM}=3$ — AM is always the larger (or equal) one for positive numbers, never the smaller.
• Reaching for Harmonic Mean or an AM–GM–HM chain. That three-way relationship uses Harmonic Progression, which is outside this syllabus; stick to the two-mean statement proved here.

Ask the tutor: want to try AM–GM on a few more optimisation-style inequalities, like minimising $2x+\tfrac{8}{x}$?