Cube root of 8 = 2 Therefore (1+a) = 2 = (1+b) = (1+c) {any other values of a, b & c that yields 8 as the answer will have at least 1 negative variable} a = 1 = b = c Therefore max of a×b×c = 1 in the domain of (1+a)(1+b)(1+c) = 8
If abcde=1 (where a,b,c,d and e are all positive real numbers) then what is the minimum value of a+b+c+d+e?
Give positive numbers a, b and c are such that ab + bc + ca + abc = 4, What is the value of 1/a+2 + 1/b+2 + 1/c+2 = ?