Show that $\int_{\mathbb{R}} |f'(t)|^2+(9t^6+18t^4)|f(t)|^2 dt\ge 3$ for functions with unit $L^2$ norm I want to show that $$g(f):=\int_{\mathbb{R}} |f'(t)|^2+(9t^6+18t^4)|f(t)|^2 dt$$ is bounded from below by $3$ for $f \in C_c^{\infty}(\mathbb{R})$ and $||f||_{L^2}=1.$ 
What is obvious is that $g$ is bounded below by $0,$ but I don't see how the $3$ comes into the game. Does anybody have an idea?
My ideas so far:
Throw away any of the terms, as they are all positive (does not sound that good to me, as it is a very bold approximation).
Use Sobolev's inequality to eliminate the derivative. 
In particular, I think we have to do something about this polynomial there. 
Use the Fourier transform (Plancherel) to turn derivatives into polynomials and vice versa.
If anything is unclear, please let me know.
 A: First of all, the lower bound you are looking for follows directly if we show that the smallest eigenvalue of the self-adjoint realization of
$$
\mathcal L=-\frac{d^2}{dt^2}+9t^6+18t^4
$$
in $L^2(\mathbb R)$ is bounded below by $3$. This fact follows by considering the Rayleigh–Ritz quotient for $\mathcal L$, since $C_c^{+\infty}$ is included in the domain of $\mathcal L$.

Statement The smallest eigenvalue of $\mathcal L$ equals $3$.

Proof
To make a rather long story short, after some failed tries to use spectral bounds (if you consider a gaussian for example, you will see that the lowest eigenvalue is less than $3.09217$, this is the reason of my comment above) , I finally came up with looking at functions in the form (this is motivated firstly by gaussians, but also by the fact that we want an even function)
$$
\psi(t)=ce^{a_2t^2+a_4t^4},
$$
where $c$ is anormalization constant. It turns out that if we let $a_2=-3/2$ and $a_4=-3/4$, i.e. consider the function
$$
\psi(t)=ce^{-\frac{3}{4}t^2(2+t^2)}
$$
then a simple differentiation shows that
$$
\mathcal L\psi=3\psi.
$$
Finally, one must argue that the eigenfunction $\psi$ corresponds to the smallest eigenvalue. But, that follows from Sturm–Liouville theory, since $\psi$ does not change sign. $\square$
As a bonus, I give you a plot of the four lowest eigenfunctions of $\mathcal L$ (plotted at the height of their corresponding eigenvalue) together with the potential (the plot was made using a slight modification of this great code).

