CQYIMO42 wrote:
Let $a, b, c$ be real numbers, $a \neq 0$. If the equation $2ax^2 + bx + c = 0$ has real root on the interval $[-1, 1]$.
Prove that
$$\min \{c, a + c + 1\} \leq \max \{|b - a + 1|, |b + a - 1|\},$$and determine the necessary and sufficient conditions of $a, b, c$ for the equality case to be achieved.
Let $t$ be the real root of the equation in interval [-1,1]: $|t|\leq 1$.
Rewriting the equation as follow; $-(a-1)t^2-bt=(a+1)t^2+c$
$$|(a-1)t+b|\cdot |t|=|(a+1)t^2+c|$$Notice that $0\leq t^2\leq 1$, so $|(a+1)t^2+c|\geq min\{c,a+1+c\}$.
Also, $-1\leq t\leq 1$. Thus, $|(a-1)t+b|\cdot |t|\leq |(a-1)t+b|\leq max\{|b+a-1|,|b-a+1|\}$.
So, we are done.