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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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sapon | Student, Undergraduate | (Level 2) Honors

Posted March 12, 2011 at 1:02 AM via web

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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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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted March 12, 2011 at 1:09 AM (Answer #1)

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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

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