Prove the Arithmetic-Geometric Mean Inequality If a_1 a_2,..., a_n are nonnegative numbers, then their arithmetic mean is frac{a_1 + a_2 + ... + a_n}{

Question

Prove the Arithmetic-Geometric Mean Inequality If

a_{1} a_{2}, \dots, a_{n}

are nonnegative numbers, then their arithmetic mean is

\frac{a_{1} + a_{2} + \dots + a_{n}}{n},

and their geometric . The avthmetic-geometic mean is

\sqrt{n} {a_{1}, a_{2}, \dots a_{n}} .

The arithmetic-geometric mean equality states that the geometric mean is always less than or equal to the arithmetic mean. In this problem we prove this in the case of two numbers andy. (a) If x and y are nonnegative and

x \leq y, t h e n x^{2} \leq y^{2} .

[Hint: First use Rule 3 of Inequalities to show that

x^{2} \leq x y and x y \leq y^{2} .

] (b) Prove the arithmetic-geometric mean inequality

\sqrt{x y} \leq \frac{x + y}{2}

Answer 1

We have to prove inequalities according the question. a) If x and y are non-negative and $x \leq y$ ,then we want to prove $x^{2} \leq y^{2} .$ Here we have given, $x \leq y$ Since, x is non-negative then, $x . x \leq x . y \Rightarrow x^{2} \leq x y$ ...(i)
Since, y is non-negative then, $x . y \leq y . y \Rightarrow x y \leq y^{2}$ ..(ii)
So,from equation (i) and (ii) we get $x^{2} \leq y^{2}$ b) For any two non-negative numbers x and y we have ${(x - y)}^{2} \geq 0$ Then, ${(x - y)}^{2} \geq 0 \Rightarrow x^{2} + y^{2} - 2 x y \Rightarrow 0 \Rightarrow x^{2} + y^{2} + 2 x y \geq 4 x y [a d d 4 x y o n ⊥ h s i d e s] \Rightarrow {(x + y)}^{2} \geq 4 x y \Rightarrow (x + y) \geq \sqrt{4 x y} \Rightarrow (x + y) \geq 2 \sqrt{x y} \Rightarrow \frac{(x + y)}{2} \geq \sqrt{x y}$
So, for any two non-negative number we have, $\sqrt{x y} \leq \frac{x + y}{2}$