# Inequality of arithmetic and geometric means

For every two nonnegative real numbers $a$ and $b$ the following inequality holds:

$$\frac{a+b}{2} \ge \sqrt{ab}.$$

The inequality above is called the inequality of arithmetic and geometric means.

The equality is valid if and only if $a = b$.

Proof.

Let’s assume that the statement is valid. Then we can write it in the following form:

$$a + b \ge 2 \sqrt{ab}$$

$$\Longleftrightarrow a – 2 \sqrt{ab} + b \ge 0$$

$$\Longleftrightarrow( \sqrt{a} – \sqrt{b}) ^2 \ge 0.$$

$( \sqrt{a} – \sqrt{b}) ^2$ is always greater than or equal to $0$, therefore, the statement is true.

Geometrical interpretation

Assume that $a>b$, without loss generality. Construct a circle with the center at point $S$ and diameter $a+b$, whereby $a= |AB|$ and $b= |BC|$. It follows that the radius is $\frac{a + b}{2}$.

Furthermore, construct the perpendicular bisector $\overline{BD}$. The length of the perpendicular bisector $\overline{BD}$ is equal  to $\sqrt{ab}$, what we can show by using the Pythagorean theorem.

The radius $\overline{SD}$ is the arithmetic mean and the altitude $\overline{BD}$ is the geometric mean.

Since the triangle $SBD$ is the right triangle, then its hypotenuse $\overline{SD}$ has the greater length than the leg $\overline{BD}$ of the same triangle.

Another geometrical approach

Assume that $a<b$. Construct the square with the length of the segment $a+b$ and right-angled triangles with legs of the length $a$ and $b$ (blue triangles). The total area of the blue triangles is $2ab$. Since the red triangles are reflections across hypotenuses of the blue ones, then the total area of the red triangles is also $2ab$. Therefore, the total area of all triangles is $4ab$.

The area of the given square is $(a+b)^2$, therefore, finally we have

$$(a+b)^2 \ge 4ab /^{\sqrt{}}$$

$$\sqrt{(a+b)^2} \ge \sqrt{4ab}$$

$$a+b \ge 2 \sqrt{ab} /:2$$

$$\frac{a+b}{2} \ge \sqrt{ab}.$$

In general, for $a_1, a_2, \cdots, a_n$, $a_i > 0$, $i = 1, 2, \cdots, n$ is valid:

$$\frac{a_1+a_2 + \cdots + a_n}{n} \ge \sqrt[n]{a_1\cdot a_2 \cdot \ldots \cdot a_n}.$$

Example 1.

If $x + y + z =1$, then $x^2 + y^2 + z^2 \ge \frac{1}{3}$. Prove it!

Solution:

After squaring the condition  $x + y + z =1$, we obtain:

$$x^2 + y^2 + z^2 + 2(xy + yz + xz) =1.$$

It follows:

$$xy + yz + xz = \frac{1 – x^2 – y^2 – z^2}{2}.$$

By using the inequality of arithmetic and geometric means on two numbers, we have:

$$\frac{x^2 + y^2}{2} \ge \sqrt{x^2 \cdot y^2} = |xy| \ge xy$$

$$\Rightarrow x^2 + y^2 \ge 2xy$$

Analogously, we obtain:

$$x^2 + z^2 \ge 2xz$$

and

$$y^2 + z^2 \ge 2 yz.$$

By adding the inequality above, we obtain:

$$x^2 + y^2 + x^2 + z^2 + y^2 + z^2 \ge 2xy + 2xz + 2 yz$$

$$\Leftrightarrow x^2 + y^2 + z^2 \ge xy + xz + yz$$

By the condition, we have $xy + yz + xz = \frac{1 – x^2 – y^2 – z^2}{2}$, that is

$$x^2 + y^2 + z^2 \ge \frac{1 – x^2 – y^2 – z^2}{2} / \cdot 2$$

$$\Leftrightarrow 2 x^2 + 2 y^2 + 2 z^2 \ge 1 – x^2 – y^2 – z^2$$

$$\Leftrightarrow 3 x^2 + 3 y^2 + 3 z ^2 \ge 1$$

$$\Leftrightarrow x^2 + y^2 + z^2 \ge \frac{1}{3}$$

Example 2.

Prove that for $a_1, a_2, a_3, a_4$ and $a_1 + a_2 + a_3 + a_4$ is valid:

$$\left ( \frac{1}{a_1} – 1\right) \cdot \left ( \frac{1}{a_2} -1 \right) \cdot \left( \frac{1}{a_3}-1 \right) \cdot \left( \frac{1}{a_4}-1 \right) \ge 81.$$

Solution:

$$\left ( \frac{1}{a_1} – 1\right) \cdot \left ( \frac{1}{a_2} -1 \right) \cdot \left( \frac{1}{a_3}-1 \right) \cdot \left( \frac{1}{a_4}-1 \right) = \left ( \frac{1 – a_1}{a_1} \right) \cdot \left ( \frac{1 – a_2}{a_2} \right) \cdot \left ( \frac{1 – a_3}{a_3} \right) \cdot \left ( \frac{1 – a_4}{a_4} \right)$$

From the condition, we have:

$$1 – a_1 = a_2 + a_3 + a_4$$

$$1 – a_2 = a_1 + a_3 + a_4$$

$$1 – a_3 = a_1 + a_2 + a_4$$

$$1 – a_4 = a_1 + a_2 + a_3$$

Now we have:

$$\left ( \frac{1}{a_1} – 1\right) \cdot \left ( \frac{1}{a_2} -1 \right) \cdot \left( \frac{1}{a_3}-1 \right) \cdot \left( \frac{1}{a_4}-1 \right)=$$

$$= \left ( \frac{a_2 + a_3 + a_4}{a_1} \right) \cdot \left ( \frac{a_1 + a_3 + a_4}{a_2} \right) \cdot \left ( \frac{a_1 + a_2 + a_4}{a_3} \right) \cdot \left ( \frac{a_1 + a_2 + a_3}{a_4} \right).$$

Now we use the inequality of arithmetic and geometric means for three numbers:

$$\frac{a_2 + a_3 + a_4}{3} \ge \sqrt[3]{a_2a_3a_4} \Leftrightarrow a_2 + a_3 + a_4 \ge 3 \sqrt[3]{a_2a_3a_4}.$$

Analogously;

$$a_1 + a_3 + a_4 \ge 3 \sqrt[3]{a_1a_3a_4}$$

$$a_1 + a_2 + a_4 \ge 3 \sqrt[3]{a_1a_2a_4}$$

$$a_1 + a_2 + a_3 \ ge 3 \sqrt[3]{a_1a_2a_3}$$

Finally we have:

$$\left ( \frac{1}{a_1} – 1\right) \cdot \left ( \frac{1}{a_2} -1 \right) \cdot \left( \frac{1}{a_3}-1 \right) \cdot \left( \frac{1}{a_4}-1 \right) =$$

$$=\left ( \frac{a_2 + a_3 + a_4}{a_1} \right) \cdot \left ( \frac{a_1 + a_3 + a_4}{a_2} \right) \cdot \left ( \frac{a_1 + a_2 + a_4}{a_3} \right) \cdot \left ( \frac{a_1 + a_2 + a_3}{a_4} \right)$$

$$\ge \frac{3 \sqrt[3]{a_2a_3a_4}}{a_1} \cdot \frac{3 \sqrt[3]{a_1a_3a_4}}{a_2} \cdot \frac{3 \sqrt[3]{a_1a_2a_4}}{a_3} \cdot \frac{3 \sqrt[3]{a_1a_2a_3}}{a_4} = \frac{3^4 \sqrt[3]{a_1^3\cdot a_2^3 \cdot a_3^3 \cdot a_4^3}}{a_1\cdot a_2 \cdot a_3 \cdot a_4} = 3^4 = 81.$$

Example 3.

Prove that $\forall a, b >0$ the following is valid:

$$2 \cdot \sqrt{a} + 3 \cdot \sqrt[3]{b} \ge 5 \cdot \sqrt[5]{ab}.$$

Solution:

We use the inequality of arithmetic and geometric means:

$$2 \cdot \sqrt{a} + 3 \cdot \sqrt[3]{b} = \sqrt{a} + \sqrt{a} + \sqrt[3]{b} + \sqrt[3]{b} + \sqrt[3]{b}$$

$$\ge 5 \cdot \sqrt[5]{\sqrt{a} \cdot \sqrt{a} \cdot \sqrt[3]{b} \cdot \sqrt[3]{b} \cdot \sqrt[3]{b}}$$

$$\ge 5 \cdot \sqrt[5]{ab}$$

Example 4.

Prove that for all real numbers $a, b, c$ is valid:

$$a^4 + b^4 + c^4 \ge abc(a + b +c).$$

Solution:

We will use the inequality of arithmetic and geometric means:

$$\frac{a^4 + b^4}{2} \ge \sqrt{a^4 \cdot b^4} \Rightarrow a^4 + b^4 \ge 2 a^2 b^2$$

Analogously;

$$b^4 + c^4 \ge 2 b^2 c^2$$

$$c^4 + a^4 \ge 2 c^2 a^2$$

Now we have:

$$2 a^4 + 2 b^4 + 2 c^4 = (a^4 + b^4) + ( b^4 + c^4) + (c^4 + a^4)$$

$$\ge 2 a^2b^2 + 2 b^2 c^2 + 2 c^2 a^2$$

$$= (a^2b^2 + b^2c^2 ) +(b^2 c^2 + c^2 a^2) + (a^2b^2 + c^2 a^2)$$

$$\ge 2ab^2 c + 2bc^2a + 2ca^2b$$

$$= 2abc( a+b +c)$$

That is,

$$a^4 +b^4 + c^4 \ge abc(a + b + c).$$