Power Inequality

Algebra Level 3

Let $a,b$ and $c$ be positive numbers such that $a+b+c = 1$ .

If the maximum value of $a^b b^c c^a$ can be expressed as $\dfrac mn$ , where $m$ and $n$ are coprime positive integers, find $m+n$ .

The answer is 4.

1 solution

Reynan Henry
Dec 26, 2016

with AM-GM $1=(a+b+c)^2\ge 3(ab+bc+ca)$ so $\frac{1}{3}\ge ab+bc+ca$ . On the other hand $ab+bc+ca = \frac{ab+bc+ca}{a+b+c}\ge \sqrt[a+b+c]{a^bb^cc^a}=a^bb^cc^a$ (this is called weighted AM-GM) so the answer is $\frac{1}{3}$

How ab + BC+ca = that thing

ankit raj - 4 years, 5 months ago

because $a+b+c = 1$ and dividing something with one is not changing anything

Reynan Henry - 4 years, 5 months ago

For completeness, you should show the equality case.

Sharky Kesa - 4 years, 5 months ago

hey, how did you bring out the first inequality by directly using AM-GM? I can't figure it out.

Dhrubajyoti Ghosh - 3 years, 10 months ago

Power Inequality

The answer is 4.

1 solution

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