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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:

    \[\begin{aligned} 2 \cdot \sqrt{a} + 3 \cdot \sqrt[3]{b} &= \sqrt{a} + \sqrt{a} + \sqrt[3]{b} + \sqrt[3]{b} + \sqrt[3]{b}\\ & \ge 5 \cdot \sqrt[5]{\sqrt{a} \cdot \sqrt{a} \cdot \sqrt[3]{b} \cdot \sqrt[3]{b} \cdot \sqrt[3]{b}}\\ & \ge 5 \cdot \sqrt[5]{ab}.\\ \end{aligned}\]

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:

    \[\begin{aligned} 2 a^4 + 2 b^4 + 2 c^4 &= (a^4 + b^4) + ( b^4 + c^4) + (c^4 + a^4) \\ & \ge 2 a^2b^2 + 2 b^2 c^2 + 2 c^2 a^2\\ & = (a^2b^2 + b^2c^2 ) +(b^2 c^2 + c^2 a^2) + (a^2b^2 + c^2 a^2) \\ & \ge 2ab^2 c + 2bc^2a + 2ca^2b \\ & = 2abc( a+b +c) \end{aligned}\]

That is,

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

 

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