$$ \sqrt{1+6x}\le(1+x)^3 $$
I understood the AM is always greater or equal to the GM thing but it was for all values greater than $0$. This doesnt look like its going to be solved the same way. Do we use Cauchy-Schwarz or the AM-GM for this? Don't forget the title for the value of $x$.
For $0\leq\alpha\leq 1$ the graph $\gamma$ of the function $b_\alpha(y):=(1+y)^\alpha$ is concave, since $$b_\alpha''(y)=\alpha(\alpha-1)(1+y)^{\alpha-2}\leq0\qquad(y\geq-1)\ .$$ It follows that $\gamma$ is for all $y\geq-1$ below its tangent at $(0,1)\in\gamma$. In this way one obtains Bernoulli's inequality $$(1+y)^\alpha\leq 1+\alpha y\qquad(y\geq-1, \ 0\leq \alpha\leq1)\ .$$
With $y:=6x$ and $\alpha:={1\over6}$ we at once obtain the claim.
Bernoulli's inequality: Let $t\ge -1$ be a real number, and $n$ a non-negative integer. Then $$ (1+t)^n\ge 1+nt. $$
Let's use $1+3x$ as a stepping stone.
Bernoulli's inequality ($n=3$, $t=x$) gives $$ (1+x)^3\ge 1+3x $$ for all $x\ge-1$.
Also ($n=2$, $t=3x$) $$ (1+3x)^2\ge 1+6x $$ whenever $3x\ge-1$ or, equivalently, when $x\ge-1/3$.
You can take the square root of the latter inequality, if both of them are non-negative (observe the range of $x$ to that end).
And then you can connect the dots.
Differentiating $RHS-LHS$ gives
$$3(x+1)^2-\frac{3}{\sqrt{6x+1}}$$
which has $0$ as its only real root, so the minimum of the difference occurs at $x=0$, where it is $0$.
Therefore, the inequality holds wherever both $LHS$ and $RHS$ are defined, which is $x\ge-\frac{1}{6}$.