Ask about inequality, A3 + B3 + C3 + 3ABC > 2 (a + b) C2 a > 0, b > 0, C > 0, a + b > C, a, B, C are not equal to each other. Who can help me to prove that? Where A3, B3, C3 is the cube of a, B, C, C2 is the square of C,

Ask about inequality, A3 + B3 + C3 + 3ABC > 2 (a + b) C2 a > 0, b > 0, C > 0, a + b > C, a, B, C are not equal to each other. Who can help me to prove that? Where A3, B3, C3 is the cube of a, B, C, C2 is the square of C,

a3+b3+c3+3abc =(a+b)(a2-ab+b2)+c(c2+3ab) >c(a2-ab+b2+c2+3ab) =c[(a+b)2+c2] >=2c(a+b)c =2(a+b)c2

Let y = x + 2 / x + 1, X ∈ (0, √ 2) ∪ (√ 2, + ∞)
1. Verification: X - √ 2 is different from Y - √ 2
2. Which one of X and Y is closer to √ 2

1. Syndrome: X - √ 2 divided by Y - √ 2, finally reduced to (x - √ 2) (x + 1) / X (1 - √ 2) + 2 - √ 2
Let the above equation be the equation, when x belongs to (0, √ 2), (x - √ 2) (x + 1) 0 [1 - √ 2]

A (x-1) / X-2 > 1 and a is not equal to 1
How to solve the inequality about x

X-2 at the denominator, not equal to 0
So (X-2) ^ 2 > 0
Multiply both sides by (X-2) ^ 2
Unequal sign does not change direction
a(x-1)(x-2)>(x-2)^2
(x-2)(ax-a-x+2)>0
(x-2)[(a-1)x-(a-2)]>0
This is to compare the size of the roots 2 and (A-2) / (A-1)
If 2 - (A-2) / (A-1) > 0, i.e. a > 1 or A1, then the solution of the inequality is
x> 2 or X