Let a, b, c, d be real numbers.

Question

(a) Prove that a² +b² +c² +d² ≥ ab + bc+cd + da.
(b) Prove that a² +b² +c²+d² ≥ 3/2ab + 1/2bc + 3/2cd+1/2da.

Answer 1

(a) To prove that a² + b² + c² + d² ≥ ab + bc + cd + da, we will use the Cauchy-Schwarz inequality.

The Cauchy-Schwarz inequality states that for any real numbers x₁, x₂, y₁, y₂, the following inequality holds:

(x₁² + x₂²)(y₁² + y₂²) ≥ (x₁y₁ + x₂y₂)²

Let's use this inequality to prove the given inequality.

Set x₁ = a, x₂ = c, y₁ = b, and y₂ = d. Then the Cauchy-Schwarz inequality becomes:

(a² + c²)(b² + d²) ≥ (ab + cd)²

Expanding both sides of the inequality, we get:

a²b² + a²d² + b²c² + c²d² ≥ a²b² + 2abcd + c²d²

Rearranging terms, we have:

a²d² + b²c² ≥ 2abcd

Now, we want to prove that a² + b² + c² + d² ≥ ab + bc + cd + da.

Notice that:

a² + b² + c² + d² = a²d² + a²b² + b²c² + c²d²

Therefore, we have:

a² + b² + c² + d² ≥ ab + bc + cd + da

This completes the proof.

(b) To prove that a² + b² + c² + d² ≥ (3/2)ab + (1/2)bc + (3/2)cd + (1/2)da, we will use the Cauchy-Schwarz inequality again.

Set x₁ = a, x₂ = √3c, y₁ = √3b, and y₂ = d. Then the Cauchy-Schwarz inequality becomes:

(a² + 3c²)(3b² + d²) ≥ (√3ab + √3cd)²

Expanding both sides of the inequality, we get:

3a²b² + a²d² + 9b²c² + 3c²d² ≥ 3ab√3cd

Rearranging terms, we have:

3a²b² + 9b²c² + a²d² + 3c²d² ≥ 3√3abcd

Notice that:

3a²b² + 9b²c² + a²d² + 3c²d² = (3/2)(2a²b² + 6b²c² + 2a²d² + 6c²d²)

From part (a), we know that 2a²b² + 6b²c² + 2a²d² + 6c²d² ≥ 4abcd

Therefore, we have:

(3/2)(2a²b² + 6b²c² + 2a²d² + 6c²d²) ≥ (3/2)(4abcd)

Simplifying, we get:

3a²b² + 9b²c² + a²d² + 3c²d² ≥ 6abcd

Now, we want to prove that a² + b² + c² + d² ≥ (3/2)ab + (1/2)bc + (3/2)cd + (1/2)da.

Notice that:

a² + b² + c² + d² = 3a²b² + 9b²c² + a²d² + 3c²d²

Therefore, we have:

a² + b² + c² + d² ≥ (3/2)ab + (1/2)bc + (3/2)cd + (1/2)da

This completes the proof.