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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.
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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
Posted by giorgiana1976 on March 12, 2011 at 1:09 AM (Answer #1)
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