Inequality in a Polynomial

Question

This is a problem I made.

Let $x^3-ax^2+bx-c$ be a polynomial with real coefficients and three real roots, all greater than $1$. Prove, that $b+c \geq 3a-5$.

Due to the discussion made (see the comments) I will post my solution (this solution does exist!) in a couple of hours

Answer 1

Since the cubic $$ (u+1)^3-a(u+1)^2+b(u+1)-c= u^3 - (a-3) u^2 + (3-2a+b)u - (c+a-b-1) $$ has all three roots positive, the coefficients $a-3$, $3-2a+b$ and $a+c-b-1$ are all positive. Hence $$ 5+b+c-3a=(c+a-b-1)+2(3-2a+b)>0. $$

Answer 2

Let $p+1$, $q+1$, $r+1$ be our roots.

Thus, $p$, $q$ and $r$ are positives and by using the Viete's theorem we need to prove that: $$\sum_{cyc}(p+1)(q+1)+\prod_{cyc}(p+1)\geq3\sum_{cyc}(p+1)-5$$ or $$pqr+2(pq+pr+qr)\geq0,$$ which is obvious.