Question

Prove: Arithmetic-Geometric Mean Inequality

If$a_1,a_2,\mathrm{.....},a_n$are non-negative numbers, then their arithmetic mean is$\frac{a_1+a_2+\mathrm{....}+a_n}{n}$, and their geometric mean is $\sqrt[n]{a_1{a}_2\mathrm{...}{a}_n}$. The arithmetic-geometric mean is always less than or equal to the arithmetic mean. In this problem we prove this in the case of two numbers$x\text{and}y$.

1. If x and y are non-negative and $x\le y,$then $x^2\le y^2$. [Hint : First use rule 3 of inequalities to show that $x^2\le xy$and $xy\le y^2$].
2. Prove the arithmetic-geometric mean inequality
   $\sqrt{xy}\le \frac{x+y}{2}$
   

Short answer

1. If$x\le y$, then$x^2\le y^2$.
2. Also, the arithmetic-geometric mean inequality is given by$\sqrt{xy}\le \frac{x+y}{2}$

Step-by-step solution

Part a. Step 1. Apply the concept of inequalities.

An inequality looks like an equation, except that in the place of the equal sign is one of the symbols, $<,>,\le ,\ge$.

Example of an inequality is $4x+7\le 19$.

An inequality is linear if each term is constant or a multiple of the variable. To solve a linear inequality, isolate the variable on one side of the inequality sign.

Rule 3: If $C>0,\text{then}A\le B\iff CA\le CB$

Rule 7: If$A\le B\text{and}B\le C$ then$A\le C$ (Inequality is transitive).

Part a. Step 2. Simplify the inequality.

Since $x\le y$, by using rule 3, multiply both the sides by x. Therefore,

$x^2\le xy-----\left(1\right)$

Also multiplying both sides of$x\le y$ by y. Hence,

$xy\le y^2-----\left(2\right)$

Part a. Step 3. Prove the inequality.

Combining the inequalities (1) and (2) by using the rule 7, the following is obtained:

$\begin{array}{c}{x}^2\le xy\le y^2\\ \Rightarrow x^2\le y^2\end{array}$

Hence proved.

Part b. Step 1. Apply the concept of inequalities.

An inequality looks like an equation, except that in the place of the equal sign is one of the symbols, $<,>,\le ,\ge$.

Example of an inequality is $4x+7\le 19$.

An inequality is linear if each term is constant or a multiple of the variable. To solve a linear inequality, isolate the variable on one side of the inequality sign.

Rule 3: If $C>0,\text{then}A\le B\iff CA\le CB$

Rule 7: If$A\le B\text{and}B\le C$ then$A\le C$ (Inequality is transitive).

Part b. Step 2. Simplify the inequality.

Prove the arithmetic-geometric mean inequality by contradiction. Hence, take $\sqrt{xy}>\frac{x+y}{2}$.

Therefore, squaring on both sides, the following is obtained:

$$\begin{gathered} \sqrt{xy}>\frac{x+y}{2} \\ \Rightarrow {\left(\sqrt{xy}\right)}^2>{\left(\frac{x+y}{2}\right)}^2 \\ \Rightarrow xy>\frac{{\left(x+y\right)}^2}{4} \\ \Rightarrow xy>\frac{x^2+y^2+2xy}{4} \end{gathered}$$

Part b. Step 3. Arrive the contradiction.

On cross multiplication the following is obtained and simplify them further:

$\begin{array}{l}\Rightarrow 4xy>x^2+y^2+2xy\\ \Rightarrow 4xy-2xy>x^2+y^2\text{(subtract}2xy)\\ \Rightarrow 2xy>x^2+y^2\\ \Rightarrow 0>x^2+y^2-2xy\text{(subtract}2xy)\end{array}$

On further simplification, it is obtained that${(x-y)}^2<0\text{as}{(x-y)}^2=x^2+2xy-y^2$

which is a contradiction as square of number can ever be negative.

Hence,$\sqrt{xy}\le \frac{x+y}{2}$