Basic inequalities proof : if a > b - c implies a > c ?

Question

I have $a,b,c \in \mathbb{R}^{+*}$ (strictly positive real numbers) and $ a > b - c$. Does it imply that : $a > c $ ? if yes hints to the proof may be helpful.

Thanks in advance

Answer 1

If $b$ and $c$ are both small, your idea works fine. E.g. $5>4-1; 5>1$ However, if they are both large (relative to $a$) and close together, the theory falls apart. For example: $5>(10,005-10,001)$, but clearly $5\not>10,001$. I'm sure there are infinitely many such counterexamples.

Answer 2

This is going to be a short answer, which could very well just be a comment, but I would like for you to remember this in the future:

Substitute $b = a + c - 1$. Then, $a > a + c - 1 - c$ if and only if $a > a - 1$ or $1 > 0$, which is true. So, $b \leqslant a + c - 1$.

But, since all of them are positive, then if $a > b - c$, one has that $a > b$. I think this is what you meant, and not $a > c$.