Proof-Theorem-Lemma Template

Definition

A function ƒ:ℝ→ℝ is convex if for all x,y∈ℝ and 0≤λ≤1

ƒ*(λ*x+(1-λ)*y)≤λ*ƒ(x)+(1-λ)*ƒ(y)

Lemma

For all real numbers x,y

(x-y)^2≥0

Proof

Expanding the square

(x-y)^2=x^2-2*x*y+y^2

Since a square of a real number is always non-negative,

(x-y)^2≥0

Lemma

For any real numbers x,y

(x+y)^2≥4*x*y

Proof

Start from the identity

(x+y)^2=x^2+2*x*y+y^2

Rewrite the expression

(x+y)^2-4*x*y=x^2-2*x*y+y^2=(x-y)^2

Using the previous lemma

(x-y)^2≥0

we obtain

(x+y)^2≥4*x*y

Theorem

For any real numbers (x_1),(x_2),…,(x_n)

((∑_i=1^n)((x_i)))^2≤n*(∑_i=1^n)((x_i)^2)

Proof

Start from the expansion

((∑_i=1^n)((x_i)))^2=(∑_i=1^n)((x_i)^2)+2*(∑_i<j^)((x_i)*(x_j))

Using the inequality

2*(x_i)*(x_j)≤(x_i)^2+(x_j)^2

we obtain

2*(∑_i<j^)((x_i)*(x_j))≤(∑_i<j^)((x_i)^2+(x_j)^2)

Substituting this into the original expression gives

((∑_i=1^n)((x_i)))^2≤n*(∑_i=1^n)((x_i)^2)

Corollary

For any real numbers (x_1),(x_2),…,(x_n)

1/n*(∑_i=1^n)((x_i)^2)≥(1/n*(∑_i=1^n)((x_i)))^2