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VISIBLE MATH｜AM–GM

AM–GM Inequality: Interactive Visual Proof

Use a right triangle inscribed in a semicircle to compare the radius $R$ with the altitude $h$, and see why the arithmetic mean is never less than the geometric mean.

Target inequality
$\dfrac{a+b}{2}\ge \sqrt{ab}$
Equality holds if and only if $a=b$

Semicircle construction

ABCDObahR$a+b$$R=\dfrac{a+b}{2}$
Move the sliders to change $a$ and $b$. Notice that $h=\sqrt{ab}$ never exceeds $R=\dfrac{a+b}{2}$.

Geometric setup

  • $AD=b$ and $DB=a$
  • $AB=a+b$
  • $R=\dfrac{a+b}{2}$

Key facts

  • An angle subtended by a semicircle is a right angle
  • The altitude to the hypotenuse satisfies $h^2=ab$
  • This follows from the similarity of the two smaller triangles
  • Therefore $h=\sqrt{ab}$

The inequality

  • The perpendicular is the shortest distance, so $h\le R$
  • $\sqrt{ab}\le \dfrac{a+b}{2}$

Equality case

  • $h=R$
  • Point C is at the top of the semicircle
  • Therefore $a=b$

When should you think of AM–GM?

Do not start by memorizing the formula. When a problem asks for an extreme value involving a sum and a product, first check whether AM–GM can reduce it to a standard form.

1

Check positivity first

The two quantities (or all quantities, in the multi-variable form) must be positive to apply the standard AM–GM inequality directly.

2

Identify what is fixed

If the problem gives, or can be rearranged into, a fixed sum or a fixed product, that is a strong signal.

3

Identify the requested extreme

Fixed product → the sum has a minimum
Fixed sum → the product has a maximum

4

Find the equality case

The extreme usually occurs when the two terms are equal: A=B. With more terms, all terms are equal.

Basic form: A＋B → AB

$A+B=S$ fixed→Maximum of $AB$
when $A=B$

This is “sum to product”: the sum is known, and the product is maximized.

Basic form: AB → A＋B

$AB=P$ fixed→Minimum of $A+B$
when $A=B$

This is “product to sum”: the product is known, and the sum is minimized.

Restriction: Standard AM–GM requires all compared terms to be positive. If a term may be zero or negative, or the domain does not guarantee positivity, you cannot apply it merely because the expression “looks like a sum and a product.”

Advanced recognition: what if coefficients hide the structure?

Do not force the original expression to match the formula. Rename the two positive terms as new quantities $A$ and $B$, then check whether their sum or product becomes fixed. A perfect-square product can make the arithmetic cleaner, but it is not required for AM–GM.

① Split into two positive termsDecide which two parts will play the roles of $A$ and $B$.
② Rename the termsSet $A=$ the first term and $B=$ the second term.
③ Check A+B or ABIf one becomes a constant, return to the basic fixed-sum or fixed-product form.
④ Use equalityAfter the structure is clear, use $A=B$ to locate the extreme and then recover the original variables.
ExampleFind the minimum of $3x+\dfrac{12}{x}$ for $x>0$.
Let $A=3x,\quad B=\dfrac{12}{x}$
Then $AB=3x\cdot\dfrac{12}{x}=36=6^2$, so the product is fixed.
Thus $A+B\ge 2\sqrt{AB}=12$, with equality when $A=B$.
From $3x=\dfrac{12}{x}$, we get $x=2$.
SimilarGiven $A,B>0$ and $2A+3B=36$, find the maximum of $AB$.
Let $x=2A,\quad y=3B$
Then $x+y=36$, so the problem becomes the basic “fixed sum → maximum product” form.
By AM–GM, $xy\le\left(\dfrac{x+y}{2}\right)^2=18^2=324$.
Since $xy=(2A)(3B)=6AB$, we have $6AB\le324$, hence $AB\le54$.
Equality occurs when $x=y$, so $2A=3B=18$. Therefore $A=9,\ B=6$, and $AB=54$.
SimilarGiven $X,Y>0$ and $XY=25$, find the minimum of $2X+3Y$.
Let $A=2X,\quad B=3Y$
Then $AB=(2X)(3Y)=6XY=150$, so the problem becomes the basic “fixed product → minimum sum” form.
By AM–GM, $A+B\ge2\sqrt{AB}=2\sqrt{150}=10\sqrt6$.
Therefore $2X+3Y\ge10\sqrt6$.
Equality occurs when $A=B$, i.e. $2X=3Y$. Together with $XY=25$, this gives $X=\dfrac{5\sqrt6}{2},\ Y=\dfrac{5\sqrt6}{3}$.
Hence the minimum of $2X+3Y$ is $\boxed{10\sqrt6}$.
Recognition tip: Making the fixed product a perfect square is a useful calculation shortcut, not a condition for AM–GM. If the product is not a perfect square, the method still works—the answer may simply contain a radical.

Algebra view: with a fixed sum, when is the product greatest?

Rewrite the same idea in algebraic form. If A+B is fixed, when is AB largest? Completing the square shows that the maximum occurs at A=B.

Proof by completing the square

Let A=x and B=S-x, where S=A+B
AB = x(S-x)
AB = -x² + Sx
AB = -(x - S/2)² + S²/4
Since -(x - S/2)² ≤ 0, we have AB ≤ S²/4

Using the current values

ABx=A
For fixed S, the parabola reaches its highest point at x=S/2, so A=B. Equivalently, for a fixed A+B, the largest possible product is (A+B)²/4, giving √AB ≤ (A+B)/2.

Going further: three or more positive numbers

The semicircle gives a direct picture for two positive numbers, but the AM–GM inequality extends to any number of positive variables. Higher-dimensional cases are harder to draw, yet the inequality remains true.

Three positive numbers

$\dfrac{a+b+c}{3}\ge \sqrt[3]{abc}$

Equality holds if and only if $a=b=c$.

$n$ positive numbers

$\dfrac{x_1+x_2+\cdots+x_n}{n}\ge \sqrt[n]{x_1x_2\cdots x_n}$

Equality holds if and only if $x_1=x_2=\cdots=x_n$.

Fixed sum and maximum product

Fixed sum $\Longrightarrow$ the product is greatest when all values are equal

For two numbers, $a=b$; for three, $a=b=c$; the same pattern continues.

Key idea: the semicircle makes the two-variable case visible. The same intuition then extends algebraically to the general AM–GM inequality.
UNDERSTAND → PRACTICE

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