Prove that a square + b square is greater than or equal to 2Ab by using the formula of sum of two numbers square

Prove that a square + b square is greater than or equal to 2Ab by using the formula of sum of two numbers square

(a-b)²≥0
a²-2ab+b²≥0
a²+b²≥2ab

It is known that a ∶ B ∶ C = 4 ∶ 3 ∶ 2, and a + 2b-4c = 4. (1) find the value of a, B, C (2) find the value of 2a-3b + C

Let a = 4x, B = 3x, C = 2x, and substitute a + 2b-4c = 4 to get x =?
So a, B, C can be solved
2a-3b + C can also be obtained
I'll just talk about the method ~ the key is to do it yourself. It's not effective to see the answer directly~

(5a-3b)(5a+3b)(25a²-9b²),(2a²-3b+4c)(2a²+3b-4c),

The original formula = (25a & # 178; - 9b & # 178;) &# 178;
=625a^4-450a²b²+81b^4
Original = [2A & # 178; - (3b-4c)] [2A & # 178; + (3b-4c)]
=(2a²)²-(3b-4c)²
=4a²-9b²+24bc-16c²

If the rational numbers a, B, C, satisfy 2a-1 + 3B + 1 + 4c-1 = 0, find the value of 2Ab + C
How to do it?

｜2a－1｜+｜3b+1｜+｜4c－1｜=0,
｜2a－1｜=0,｜3b+1｜=0,｜,｜4c－1｜=0,
a=0.5 b=-1/3 c=0.25
2ab+c=-1/12