complex AM GM inequality, probably based on weighted means

Question

I came across a question where i had to find the maximum value of a²b³c², and was just provided with the value of a+b+c which was 3 While trying to solve this i felt that i could use weighted means, which gave the maximum value of a²b³c² as [2(a+b+c)/7+b/7]^7 I replaced a+b+c with 3, but was still left with 'b', how do i eliminate that?

Answer 1

You could apply AM-GM to the values $$\left\{a, a, \frac23b, \frac23b, \frac23b, c, c\right\}$$ to get $$\frac{a + a + \frac23b + \frac23b + \frac23b + c + c}{7} \ge \sqrt[7]{a \cdot a \cdot \frac23 b \cdot \frac23 b \cdot \frac23b \cdot c \cdot c}$$ which simplifies to $$\frac27(a+b+c) \ge \sqrt[7]{\frac{8}{27}a^2b^3c^2}.$$ Since $a+b+c=3$, this tells us that $$a^2 b^3 c^2 \le \frac{27}{8} \left(\frac27 \cdot 3\right)^7 = \frac{2^4\cdot 3^{10}}{7^7}$$ with equality only if $a = \frac23 b = c$, which happens when $(a,b,c) = (\frac67, \frac97, \frac67)$.

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I guess the weighted AM-GM version of this would be to say that $$\frac27 a + \frac37\left(\tfrac23b\right) + \frac27c \ge a^{2/7} \cdot \left(\tfrac23b\right)^{3/7} \cdot c^{2/7}$$ and from there proceed as above.