The second order differential equation $\Delta v+2v\ln v+(\lambda+1)v=0$ is not difficult to solve in one dimensional case, and this is what the authors of the paper need to proceed by induction. If you have a second order ODE of the type $F(v'',v)=0$ then assuming that $v'(t)=w(v(t))$ for some function $w$, we obtain $v''(t)=w'(v(t))v'(t)=w'(v(t))w(v(t))$. Thus the initial ODE takes the form $F(w'(s)w(s),s)=0$ which is the first order ODE and then you try to solve it. 

The way I understand how to deal with functional inequalities (and in particular with Log-Sobolev inequality) goes like this (besides of variational calculus approach when it happens to be difficult). First lets make the following observation:

1) The inequality holds for **all** test functions (i.e., the smooth, compactly supported nonnegative functions). 

This is a strict requirement. It should already give you a condition on *an integrand with whom you compose test functions*. For example, if the inequality $\int_{0}^{1}B(\varphi(t))dt\leq B\left(\int_{0}^{1}\varphi(t)dt\right)$ holds for **all** test functions $\varphi$ then this is equivalent to the fact that $B(x)$ is a convex function. You may argue in the similar way for the Log-Sobolev inequality:

2) Let $g=f^{2}$ then the Log-Sobolev inequality can be rewritten as follows  
$$
\int_{\mathbb{R}^{n}} M(g,\|\nabla g\|)d\mu \leq M\left( \int_{\mathbb{R}^{n}}gd\mu,0 \right) \quad \text{for all} \quad  g> 0, \quad g\in C^{\infty}_{c}(\mathbb{R}^{n}) \quad(1)
$$
 where $M(x,y)=x\ln x - \frac{y^{2}}{2x}$ (since $d\mu$ is the standard Gaussian measure we have that $c=1$ in the inequality). 

Now we expect that (1) should mimic to some *concavity* type condition on $M(x,y)$. A subtle observation is that (1) is always implied by the following *concavity type* condition:  
\\[M_{y} \leq 0 \quad \text{and} \quad \begin{pmatrix} M_{xx}+\frac{M_{y}}{y} & M_{xy} \\
M_{xy} & M_{yy} \end{pmatrix} \leq 0. \quad (2)\\]

Clearly $M(x,y)=x\ln x - \frac{y^{2}}{2x}$ satisfies this condition therefore the Log-Sobolev inequality holds. Also Bobkov's inequality can be recovered  in the same way (and many others).

Unfortunately (1) does not imply (2), so it is not *if and only if* condition. If somebody wants *if and only if*  condition then: Let $n\geq 2$. (2) holds if and only if $P_{t} M(f,\|\nabla f\|)\leq M(P_{t}f, \| \nabla P_{t} f\|)$
for all test functions $f$ and for all $t\geq 0$, where $P_{t}$ is the Ornstein--Uhlenbeck semigroup
$$
P_{t} f (x) = \int_{\mathbb{R}^{n}}f(xe^{-t}+y\sqrt{1-e^{-2t}})d\mu(y).
$$
And then (1) follows by taking $t\to \infty$. 

After reading Ryan O'Donnell's answer I am wondering (but I have never thought about this) whether (2) implies *two-point inequality* i.e., for Boolean functions and the fact that you can proceed by induction to use CLT:

$$
\frac{1}{2}\left[ M\left( a,\left|\frac{a-b}{2} \right|\right)+ M\left( b,\left|\frac{a-b}{2} \right|\right)\right] \leq M\left( \frac{a+b}{2},0\right)?
$$
One has to write some Taylor's expansion and check it.