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