Suppose one is given, $10 = a + b + c$, where $a, b, c \in R^{+}$, and is asked to find the maximum value of $a^2 b^3 c^5$
To find the maximum value, we can easily use the $AM>=GM$ inequality by first breaking the sum as:
$10 = a + b + c = 2 \frac{a}{2} + 3 \frac{b}{3} + 5 \frac{c}{5} = \frac{a}{2} +\frac{a}{2} +\frac{b}{3} +\frac{b}{3} +\frac{b}{3} + \frac{c}{5} +\frac{c}{5} +\frac{c}{5} +\frac{c}{5} +\frac{c}{5}$
Applying $AM>=GM$ to these 10 quantities,
$\frac{\frac{a}{2} +\frac{a}{2} +\frac{b}{3} +\frac{b}{3} +\frac{b}{3} + \frac{c}{5} +\frac{c}{5} +\frac{c}{5} +\frac{c}{5} +\frac{c}{5}}{10} \geq (\frac{a^2 b^3 c^5}{2^2 3^3 5^5})^{\frac{1}{10}}$
which gives on solving,
$a^2 b^3 c^5 \leq 2^2 3^3 5^5 = 337500$
which is the correct answer (I verified this method on 2 variables by plotting them on a graph and this indeed is correct).
However, another possibility could be splitting the sum as,
$10 = \frac{5(a+b+c)}{5}$
or, $4a + a + 2b + 2b + b + c + c + c + c + c = 50$
Again, applying $AM>=GM$ to these 10 quantities,
$\frac{4a + a + 2b + 2b + b + c + c + c + c + c}{10} \geq (16a^2 b^3c^5)^{\frac{1}{10}}$
which gives on solving,
$a^2 b^3 c^5 \leq \frac{5^{10}}{16} = 610351.562$
which is incorrect. The values can never attain this product if they are confined to positive Reals. We could split the sum in the second method in any other way too - the answer would still be more than the actual maxima. The answer is correct from the second method only when the powers on $a,b,c$ are equal.
My Question about this is, why? Why does the second method return incorrect results? What is happening here? Better off, what is my mistake?
It turns out that the 2nd method is only slightly incorrect.
The equality in $AM >=GM$ holds only when all the numbers are equal. Thus,
$\frac{\frac{a}{2} +\frac{a}{2} +\frac{b}{3} +\frac{b}{3} +\frac{b}{3} + \frac{c}{5} +\frac{c}{5} +\frac{c}{5} +\frac{c}{5} +\frac{c}{5}}{10} = (\frac{a^2 b^3 c^5}{2^2 3^3 5^5})^{\frac{1}{10}}$
only when $\frac{a}{2}=\frac{b}{3} =\frac{c}{5}$ which is possible for the 3 positive reals, $a,b,c$.
However, in the second case,
$\frac{4a + a + 2b + 2b + b + c + c + c + c + c}{10} \neq (16a^2 b^3c^5)^{\frac{1}{10}}$
for any combination of positive reals $a,b,c$ because the equality requires that $4a = a = 2b = b = c$ which holds only when $a = b = c = 0$ which is not allowed.
Thus for the second case, we must write,
$\frac{4a + a + 2b + 2b + b + c + c + c + c + c}{10} > (16a^2 b^3c^5)^{\frac{1}{10}}$
which does hold true, as I explained in the question. However, this does not give the maximum value and thus not the right answer.