Prove the identity c*a^2*b^3=1 if inequality 2*a^x+3*b^x+c^x>=6, for a,b,c positive.

We'll create the function f(x) = 2*a^x+3*b^x+c^x and we
know from enunciation that f(x)>=6.

We notice that
for x = 0, we'll get:

f(0) = 2*a^0+3*b^0+c^0 = 2 + 3 + 1 =
6

Since f(x) is increasing, being a sum of increasing
functions, we'll conclude that x = 0 is a minimum
point.

We'll calculate the 1st derivative of the function
f(x):

f'(x) = 2*a^x*ln a + 3*b^x*ln b + c^x*ln
c

Based on Fermat's theorem, we'll
have:

f'(0) = 0 if and only if 2ln a + 3ln b + ln c =
0

We'll apply the power rule of the
logarithms:

ln a^2 + ln b^3 + ln c =
0

We'll apply the product rule of
logarithms:

ln a^2 + ln b^3 + ln c = ln
(a^2*b^3*c)

ln (a^2*b^3*c) =
0

We'll take
anti-logarithms:

a^2*b^3*c =
e^0

a^2*b^3*c =
1