Let a, b, c, d be real numbers.
(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.
(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.
To prove both (a) and (b), we can use the Cauchy-Schwarz inequality. The Cauchy-Schwarz inequality states that for any real numbers a1, a2, ..., an and b1, b2, ..., bn, the following inequality holds:
(a1^2 + a2^2 + ... + an^2)(b1^2 + b2^2 + ... + bn^2) ≥ (a1b1 + a2b2 + ... + anbn)^2
Now let's proceed with the proofs step by step:
(a) Prove that a² + b² + c² + d² ≥ ab + bc + cd + da:
Step 1: Rewrite the inequality using the Cauchy-Schwarz inequality:
(a^2 + b^2)(1^2 + 0^2) + (b^2 + c^2)(0^2 + 1^2) + (c^2 + d^2)(1^2 + 0^2) + (d^2 + a^2)(0^2 + 1^2) ≥ (a*1 + b*0 + c*0 + d*1)^2
Step 2: Simplify:
(a^2 + b^2)(1 + 0) + (b^2 + c^2)(0 + 1) + (c^2 + d^2)(1 + 0) + (d^2 + a^2)(0 + 1) ≥ (a + d)^2
Step 3: Expand and simplify:
2(a^2 + b^2 + c^2 + d^2) ≥ (a + d)^2
Step 4: Expand and rearrange:
2(a^2 + b^2 + c^2 + d^2) ≥ a^2 + 2ad + d^2
Step 5: Simplify:
a^2 + b^2 + c^2 + d^2 ≥ ab + ad + da + bd + cd + d^2
Step 6: Rearrange and simplify:
a^2 + b^2 + c^2 + d^2 ≥ ab + bc + cd + da
Therefore, we have proved that a² + b² + c² + d² ≥ ab + bc + cd + da.
(b) Prove that a² + b² + c² + d² ≥ (3/2)ab + (1/2)bc + (3/2)cd + (1/2)da:
Step 1: Multiply both sides of the previous inequality by (1/2):
(1/2)(a^2 + b^2 + c^2 + d^2) ≥ (1/2)(ab + bc + cd + da)
Step 2: Distribute the (1/2) on the right side:
(a^2 + b^2 + c^2 + d^2)/2 ≥ (ab + bc + cd + da)/2
Step 3: Rearrange and simplify:
(a^2 + b^2 + c^2 + d^2)/2 ≥ (ab + da)/2 + (bc + cd)/2
Step 4: Rearrange and simplify:
(a^2 + b^2 + c^2 + d^2)/2 ≥ (3ab + da)/2 + (bc + 3cd)/2
Step 5: Simplify the denominators:
(a^2 + b^2 + c^2 + d^2)/2 ≥ (3ab + da + bc + 3cd)/2
Step 6: Simplify:
a^2 + b^2 + c^2 + d^2 ≥ 3ab + da + bc + 3cd
Therefore, we have proved that a² + b² + c² + d² ≥ (3/2)ab + (1/2)bc + (3/2)cd + (1/2)da.