Prove that $f(x)\ge 3e^{2x}-2e^{3x}$ for every $x\ge 0$. 
Question: Suppose $f:\mathbb{R}\to\mathbb{R}$ is a twice differentiable function with $f(0)=1$, $f'(0)=0$ and satisfies $f''(x)-5f'(x)+6f(x)\ge 0$ for every $x\ge 0$. Prove that $f(x)\ge 3e^{2x}-2e^{3x}$ for every $x\ge 0$. 

My approach: Let $h:\mathbb{R}\to\mathbb{R}$ be such that $h(x)=f(x)-3e^{2x}+2e^{3x}, \forall x\in\mathbb{R}.$ Thus $h$ is also a twice differentiable function with $$h'(x)=f'(x)-6e^{2x}+6e^{3x}, \forall x\in\mathbb{R}, \text{ and }\\h''(x)=f''(x)-12e^{2x}+18e^{3x}, \forall x\in\mathbb{R}.$$
Also observe that $$h''(x)-5h'(x)+6h(x)=f''(x)-5f'(x)+6f(x), \forall x\in\mathbb{R}.$$ 
Thus we have $$h''(x)-5h'(x)+6h(x)\ge 0, \forall x\ge 0.$$
Now for the sake of contradiction, let us assume that $\exists a>0,$ such that $$f(a)<3e^{2a}-2e^{3a}\implies f(a)-3e^{2a}+2e^{2a}<0\implies h(a)<0.$$ 
Note that $h(0)=0$. Thus, by applying MVT to the function $h$ on the interval $[0,a]$, we can conclude that $\exists c\in(0,a)$, such that $$h'(c)=\frac{h(a)-h(0)}{a-0}=\frac{h(a)}{a}\implies h'(c)<0.$$
Again, note that $h'(0)=0$. Thus by applying MVT to the function $h'$ on the interval $[0,c]$, we can conclude that $\exists c_1\in(0,c)$, such that $$h''(c_1)=\frac{h'(c)-h'(0)}{c-0}=\frac{h'(c)}{c}\implies h''(c_1)<0.$$
Thus we have $f''(c_1)-12e^{2c_1}+18e^{3c_1}<0\implies f''(c_1)<12e^{2c_1}-18e^{3c_1}<0.$ 
As one can see, I am trying to prove it using "proof by contradiction". So, is there any way to proceed on these lines, or is there some alternative way to prove?
 A: Separate the terms as $\ {f^{\prime\prime}}-3{f^{\prime}}-2(f^{\prime} - 3f ) \ge 0$
now  multiplying  by $ e^{-2x}$  and  integrating  from  0  to   x  we  get  $f^{\prime} \ - \ 3f \ge -3e^{2x}$
again  multiplying  by  $ e^{-3x}$  and  integrating  from  0  to  x
we get the desired result
A: I have no idea about following your proof by contradiction. However, I have another proof.
This problem has the same idea as a simple problem: Let $f :\mathbb{R} \to \mathbb{R}$ be continuously differentiable such that $f(0)=1$ and $f(x)-f'(x) \geq 0$ for every $x \geq 0$. Prove that $f(x) \leq e^x$ for any $x \geq 0$.
The idea is considering new function $F(x)=e^{-x}f(x)$ then $F'(x)=e^{-x}(f'(x)-f(x)) \leq 0$ for any $x \geq 0$. Then $F(x) \leq F(0)=1$ for any $x \geq 0$, which give you $f(x) \leq e^x$ for any $x \geq 0$.
Back to your problem, we use the same method.


*

*Note that $0\leq f"(x)-5f'(x)+6f(x)= f"(x)-3f'(x) + 2(3f(x)-f'(x))$. Define $g(x)=3f(x)-f'(x)$ then $g'(x) \leq 2g(x)$ and $g(0)=3f(0)-f'(0)=3$. Use the same argument with $G(x)=e^{-2x}g(x)$ we get $g(x) \leq 3 e^{2x}$.

*$g(x) \leq 3 e^{2x} \iff 3(f(x)-3e^{2x})\leq f'(x) -6e^{2x}$. Define $h(x)=f(x)-3e^{2x}$ then $3h(x) \leq h'(x)$ and $h(0)=-2$. Use the same argument with $H(x)=e^{-3x}g(x)$ we get $h(x)=f(x)-3e^{2x} \geq -2 e^{3x}$, which gives the result.
A: When you want to compare two "solutions" of an two order EDO, it is often useful to consider the Wronskien. To avoid problem of division by 0, let's consider :
$$u = f + 2 e^{3x}$$
and 
$$v = 3 e^{2x}$$
Define the Wronskien between $u$ and $v$ by :
$$W = uv' - vu'$$
By assumption :
$$W' = uv'' - vu'' \leq u (5v' - 6v) - v (5u' - 6u) = 5 W$$
But, 
$$W(0) = 0$$
By multiplying by $e^x$ and integrating, we obtain :
$$W \leq 0$$
Wich implies :
$$\frac{uv' - vu'}{v^2} = -(\frac{u}{v})' \leq 0$$
Which gives the results :
$$u \geq \frac{u(0)}{v(0)}v = v$$
