AM GM inequality states that for real positive numbers x,y
$$x + y \geq 2(xy)^{1/2}$$
You would get the least value for x = y
We can also write this as,
$$\frac x3 + \frac x3 + \frac x3 + y \geq 4\left((x^3)\cdot(y)\cdot\left(\frac{1}{27}\right)\right)^\frac14$$
You would get the least value for $x = 3y$
Why is this contradiction arising?
Also for a given set of variables being added together, how can we find the most accurate inequality using AM GM?
On
I'm not quite sure what you want to say in your question, but I would like to mention some points.
We consider $x$ and $y$ are non-negative.
$$x+y\ge 2\sqrt {xy}$$
That is correct. We say that the equality occurs iff (or if and only if) when $x=y$. We doesn't say $\min \{x+y\}=2\sqrt {xy}$. The argument is simple: Because, $xy$ is not a constant.
$$\frac x3+\frac x3+\frac x3+y≥4\sqrt [4]{\frac {x^3y}{27}}$$
The same can be said for the second inequality. We say that the equality occurs iff when $x=3y$. We doesn't say $\min \{x+y\}=4\sqrt [4]{\frac {x^3y}{27}}$. Because, $\frac {x^3y}{27}$ is not a constant.
In other words, the inequalities $$x+y≥ 2\sqrt {xy}$$ or $$x+y\ge 4\sqrt [4]{\frac {x^3y}{27}}$$ are both valid and true.
Which inequality to use will depend on what is intended.
Finally, consider the following exact example:
Let, $x\ge 0$ then what is the minimum value of $\dfrac 2x +x^2$ ?
If you apply the AM-GM inequality directly, you will get
$$\frac 2x +x^2\ge 2\sqrt {2x}$$
This inequality is always true and equality occurs iff $x=\sqrt [3]{2}$, however the right-hand side (RHS) is not a constant.
Now, let's apply the AM-GM inequality in a different way:
$$ \begin{aligned}\frac 2x+x^2&=\frac 1x+\frac 1x +x^2\\ &\ge 3\sqrt[3]{\frac 1{x^2}\times x^2}\\ &=3.\end{aligned} $$
You can observe that the result we want to achieve is exactly this and equality occurs iff when $x=1$.
Let $x,y \ge 0$.
It is true that $x + y \ge 2 (xy)^{1/2}$, and that equality holds when $x=y$.
It is also true that $x + y = \frac13x + \frac13x + \frac13 x + y \ge 4((\frac13x)^3 y)^{1/4} = \frac{4}{3^{3/4}} (x^3y)^{1/4}$, and that equality holds when $\frac13x = y$.
These do not contradict each other, and neither of them is necessarily a minimum value of $x+y$.
In some applications of $x + y \ge 2 (xy)^{1/2}$, you assume that the product $xy$ is constant. In that case and in that case only, you can say that $x+y$ is minimized when $x=y = \sqrt{xy}$.
Similarly, if the product $x^3y$ is held constant, we can use the second version of AM-GM to conclude that $x+y$ is minimized when $\frac13x = y = (\frac x3)^{3/4} y^{1/4} = \frac1{3^{3/4}} \cdot (x^3y)^{1/4}$.
These still do not contradict each other: with different conditions on $x$ and $y$, $x+y$ is minimized at different points.