Basic approximation algorithms understanding Question:

Suppose we have 2 algorithms $Alg1$ and $Alg2$ for the same minimization problem. We know that $Alg1$ is a $2$-approximation algorithm and $Alg2$ a $4$-approximation algorithm.

Is the following statement true?

There must be an input $I$ such that $Alg2(I) \geq 2 \cdot Alg1(I)$. 

My answer:

I think the statement is false:

In order to prove that the statement is false it is sufficient to find one case where $Alg1$ is a $2$-approximation and $Alg2$ is a 4 approximation and that there is no input which satisfies the inequality from the statement.

Does my approach make sense? Is there some other approach or thought process which proves statements of this type true/false that I'm missing?

Note:

$Alg$ is a $\rho$-approximation algorithm for a minimization problem $\iff$ $\forall I.(OPT(I) \leq Alg(I) \leq \rho \cdot OPT(I))$ where $\rho &gt; 1$ and $OPT(I)$ denotes the optimal solution to the minimization problem for the input $I$.
According to your definition (which is the standard definition), an algorithm solving a certain minimization problem is an $\alpha$-approximation algorithm for any $\alpha \geq 1$.

Consider a minimization problem solvable in polynomial time, for example minimum cut. Let Alg1 and Alg2 be algorithms solving this problem exactly. Alg1 is a 2-approximation algorithm and Alg2 is a 4-approximation algorithm, yet there is no input on which Alg2 performs worse than Alg1.