Proving $2P(x)>P'(x)$ with Continuous Third Derivative of $P(x)$

[anemone]

Let $P$ be a real function with a continuous third derivative such that $P(x),\,P'(x),\,P''(x),\,P'''(x)$ are greater than zero for all $x$.

Suppose that $P(x)>P'''(x)$ for all $x$, prove that $2P(x)>P'(x)$ for all $x$.

 

[anemone]

Solution proposed by other:

Spoiler

For simplicity, we will show $2P(0)>P'(0)$. Applying this result to $P(x+c)$ shows that $2P(c)>P'(c)$ for all $c$.

Since $P'(x)$ is positive, $P$ is an increasing function. Thus for $x\le 0$, $P'''(x)< P(x)\le P(0)$. Integrating $P'''(x)\le P(0)$ from $x$ to 0 gives $P''(x)\le P''(0)+P(0)x$ for $x\le 0$, and integrating again gives the second inequality in $0<P'(x)\le P'(0)+P''(0)x+P(0)\dfrac{x^2}{2}$. Thus the polynomial $P'(0)+P''(0)x+P(0)\dfrac{x^2}{2}$ has no negative zeros. It also has no nonnegative zeros (because all its coefficients are positive). Therefore its discriminant $P''(0)^2-2P(0)P'(0)$ must be negative.

In a similar vein, since $P'''$ is positive, $P''$ is increasing. Thus for $x\le 0$, $P''(x)\le P''(0)$, so $P'(x)\le P'(0)+P''(0)x$ and $0<P(x)\le P(0)+P'(0)x+P''(0)\dfrac{x^2}{2}$.

Again, the discriminant of the quadratic must be negative: $P'(0)^2-2P(0)P''(0)<0$.

Combining these conclusion, we obtain $P'(0)^4<4P(0)^2P''(0)^2<8P(0)^3P'(0)$, which implies $2P(x)>P'(x)$ for all $x$..