Prove inequality: A4 + B4 + C4 ≥ a2b2 + b2c2 + c2a2 ≥ ABC (a + B + C)

Prove inequality: A4 + B4 + C4 ≥ a2b2 + b2c2 + c2a2 ≥ ABC (a + B + C)

It is proved that: (a 4 + B 4 ≥ 2 a 2 B 2, B 4 + C 4 ≥ 2 B 2 C 2, C 4 + a 4 ≥ 2 B 2 C 2, C 4 + a 4 + a 4 ≥ 2 a 2 2 C 2 2 ᙠ 2 (a 4 + B 4 + C 4) ≥ 2 (a 2 B 2 + B 2 C 2 + a 2 C 2) that is a 4 + B 4 + C 4 ≥ a B 2 B 2 + B 2 C 2 + a 2 B 2 C 2 a B 2 B 2 C 2 ≥ a B 2 b 2 C 2; B 2 C 2 + a 2 C 2 2 2 a 2 B 2 B 2 B 2 C 2 a 2 B 2 B 2 C 2 a 2 (a 2 B 2 B 2 B 2 B 2 C 2 + a 2 C 2 C 2 C 2 C 2) ≥ 2 (a 2 B 2 B and

Let ABC belong to R proof | a ^ 2 + B ^ 2 - √ a ^ 2 + C ^ 2|

Triangle, the difference between the two sides is less than the third side
Let B (a, b), C (a, c), o be the origin
Then ob = √ [a ^ 2 + B ^ 2], OC = √ [a ^ 2 + C ^ 2], BC = | b-c|
When OBC forms a triangle, | ob-oc | BC, that is √ a ^ 2 + B ^ 2 - √ a ^ 2 + C ^ 2|

2 (a ^ 3 + B ^ 3 + C ^ 3) "a ^ 2 (B + C) + B ^ 2 (a + C) + C ^ 2 (B + a), which is proved by sorting inequality ABC is a positive number 2 (a ^ 3 + B ^ 3 + C ^ 3) (a ^ 2) (B + C) + (b ^ 2) (a + C) + (C ^ 2) (B + a), which is proved by the sort inequality

Let's say: 00

Prove inequality: A.B.C ∈ R, a ^ 4 + B ^ 4 + C ^ 4 ≥ ABC (a + B + C)

a^4+b^4>=2a^2*b^2
a^4+c^4>=2a^2*c^2
2a^4+b^4+c^4>=4a^2*bc
Similarly, 2b ^ 4 + C ^ 4 + A ^ 4 > = 4AB ^ 2 * C
2c^4+a^4+b^4>=4abc^2
Add up
4a^4+4b^4+4c^4>=4a^2*bc+4ab^2*c+4abc^2
A ^ 4 + B ^ 4 + C ^ 4 > = ABC (a + B + C)
The equal sign is obtained when a = b = C

Proof of mathematical inequality Let x ≥ 1, prove 1 + X + x ^ 2 + +X ^ 2n ≥ 2 (n + 1) x ^ n!

First of all, you have to write the wrong formula itself,
Let n = 1, x = 1, left = 3 = 2x ^ n
x^(2n-1）+x>=2x^n
...
x^(n+1)+x^(n-1)>=2x^n
x^n=x^n
You can get it by adding both sides together
1+x+x^2+…… +x^2n≥ （2n+1)x^n

x>a^2+b^2 Then x > 2Ab

prove:
From (a + b) ^ 2 > 0
A ^ 2 + B ^ 2 > 2Ab is obtained
From x > A ^ 2 + B ^ 2
X > 2Ab

Urgent, prove inequality It is proved that a > 0, b > 0, C > 0, a + B + C = 1, and ab + BC + AC ≤ 1 / 3

It is proved that (a-b) ^ 2 + (B-C) ^ 2 + (C-A) ^ 2 ≥ 0,
A ^ 2 + B ^ 2 + C ^ 2 ≥ AB + BC + Ca is obtained
So (a + B + C) ^ 2 = (a ^ 2 + B ^ 2 + C ^ 2) + 2Ab + 2BC + 2ca ≥ (AB + BC + Ca) + 2Ab + 2BC + 2ca = 3 (AB + BC + Ca)
If a + B + C = 1, we can get 1 ≥ 3 (AB + BC + Ca)
Therefore: ab + BC + AC ≤ 1 / 3, if and only if a = b = C, the equal sign holds!
The conditions of ps.a > 0, b > 0, C > 0 are redundant!

Proof of mathematical inequality It is known that a > 0. B > 0 and a + B = 1 are proved; (a + 1 / a) * (B + 1 / b) > = 25 / 4

The inequality you want to prove is 4 (a ^ 2 + 1) (b ^ 2 + 1) > = 25ab expansion, which is: 4A ^ 2B ^ 2 + 4A ^ 2 + 4B ^ 2 + 4 > = 25ab, because a + B = 1, so (a + b) ^ 2 = 1, we put forward a ^ 2 + B ^ 2 = 1-2ab, and the inequality to be proved becomes 4A ^ 2B ^ 2-33ab + 8 > = 0. Let AB = x, solve 4x ^ 2-33x + 8 = 0 and get x = 1 / 4 or 8

a> B > C to prove a ^ 2B + B ^ 2C + C ^ 2A > AB ^ 2 + BC ^ 2 + Ca ^ 2 As the title

a^2b+b^2c+c^2a-ab^2-bc^2-ca^2
=a^2(b-c)+a(c^2-b^2)+bc(b-c)
=a^2(b-c)-(ab+ac)(b-c)+bc(b-c)
=(b-c)(a^2-ac-ab+bc)
=(b-c)[a(a-c)-b(a-c)]
=(b-c)(a-b)(a-c)
Because a > b > C,
So B-C > 0, A-B > 0, a-c > 0,
So (B-C) (a-b) (A-C) > 0,
That is, a ^ 2B + B ^ 2C + C ^ 2a-ab ^ 2-bc ^ 2-Ca ^ 2 > 0,
So a ^ 2B + B ^ 2C + C ^ 2A > AB ^ 2 + BC ^ 2 + Ca ^ 2

Given that a, B and C are positive numbers, we prove that 2 (A3 + B3 + C3) ≥ A2 (B + C) + B2 (a + C) + C2 (a + b)

Proof: first prove: A3 + B3 ≥ A2B + AB2,
∵（a3+b3）-（a2b+ab2）
=a2（a-b）-b2（a-b）
=（a2-b2）（a-b）
=（a+b）（a-b）2
≥0，
If A3 + B3 ≥ A2B + AB2, the condition of equal sign is a = B,
Similarly, A3 + B3 ≥ A2B + AB2,
a3+c3≥a2c+ac2，
b3+c3≥b2c+bc2
Add the three formulas to get:
2（a3+b3+c3）≥a2（b+c）+b2（a+c）+c2（a+b），
The condition of equal sign is a = b = C,
∴2（a3+b3+c3）≥a2（b+c）+b2（a+c）+c2（a+b）．