AM GM inequality

Question

For the above question, I remember there used to be an approach called "similarity" or "covergence" in my local language, not sure what it is called generally. Basically, it is a fast paced method for solving such questions where you have to find a minimum value.

Since, AM = GM if and only if every number in the list is the same (in which case all are equal). We make every term same, and then a get an integral value 'K' for the expression. Since our assumption was that they are equal, but in actual they are not, so we conclude that the expression must be > K.

Now, I can't seem to use that method effectively over here. Kindly help me with it please, I am trying to solve it without going the long winded approach of writing down AM >= GM terms, and solving for it.

Answer 1

Let $f(x)=\frac{x^2+3x+1}{x}.$ Then $ f'(x)=\frac{x^2-1}{x^2}.$ So $ f'(x)$ attains minimum at $x=1$ (x is +ve ) (one can check the sign of $f''(1)$ for completeness of the answer) and $f(1)=5.$ The given function is $g(xyz)= 2 f(x). 2f(y).2f(z).$ Therefore the minimum value of $g(xyz)$ is $2.5.2.5.2.5=1000.$

Answer 2

Just converting @dxiv's comment to an answer, for completeness.

We have $$\frac{(2x^2 + 6x + 2)(2y^2 + 6y + 2)(2z^2 + 6z + 2)}{xyz} = \left(2x + \frac{2}{x} + 6 \right)\left(2y + \frac{2}{y} + 6 \right)\left(2z + \frac{2}{z} + 6 \right)$$ and $$2x + \frac{2}{x} + 6 \ge 6 + 4 = 10$$ by the AM-GM inequality applied to $2x$ and $\frac2x$. The minimum is attained at $x = 1$. Therefore, $$\left(2x + \frac{2}{x} + 6 \right)\left(2y + \frac{2}{y} + 6 \right)\left(2z + \frac{2}{z} + 6 \right) \ge 1000$$ and the minimum is attained at $x = y = z= 1$.