Given that real numbers a, B, C satisfy a + C = 6-4b + 3B ^ 2, C-A = 4-4b + B ^ 2, then the size relation of ABC is

Given that real numbers a, B, C satisfy a + C = 6-4b + 3B ^ 2, C-A = 4-4b + B ^ 2, then the size relation of ABC is

The sum of the two formulas leads to
2c=10+4b^2=》c=2b^2+5
Subtraction
2a=2+8b+2b^2=》a=b^2+4b+1
Calculation of C-A
be
2b^2+5-(b^2+4b+1)=>
(b-2) ^ 2 = 0
So c > = a
The other is the same way, calculate the value after subtraction, and then compare it with 0

Elden Ring Premiere

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