Given 2a-3b-4c = 4, find 4 ^ A / 8 ^ b * (1 / 16) ^ C-4 =?

Given 2a-3b-4c = 4, find 4 ^ A / 8 ^ b * (1 / 16) ^ C-4 =?

4^a/8^b*(1/16)^c-4
=2^2a/2^3b*2^(-4c)-4
=2^(2a-3b-4c)-4
=2^4-4
=16-4
=12

The minimum value of 2A + 3B + 4C is known: a + B + C = 1, a > b > C > 0

The two dimensional form of Cauchy inequality (a ^ 2 + B ^ 2) (C ^ 2 + D ^ 2) ≥ (AC + BD) ^ 2 generalizes (A1 · B1 + A2 · B2 + a3 · B3 +... + an · BN) ^ 2 ≤ (A1 ^ 2) + (A2 ^ 2) + (A3 ^ 2) +... + (an ^ 2)) ((B1 ^ 2) + (B2 ^ 2) + (B3 ^ 2) +... (BN ^ 2)) so: (2a + 3B + 4C) * (1 / 2 + 1 / 3 + 1 / 4) ≥ (radical 2A * radical 1 / 2)

Given 2a-3b-4c = 5, find the value of 4 ^ A / 8 ^ b * (1 / 16) ^ C

4^a/8^b*(1/16)^c=2^2a/8^3b×16^-c=2^2a/8^3b×2^-4c=2^2a-3b-4c=2^5=32