$\text{ If }f\left(x\right)=a{x}^2+bx+c\text{ and }f(-1)\ge -4,f\left(1\right)\le 0\text{ and }f\left(3\right)\ge 5\text{ then the least value of }a\text{ is }$

$\begin{array}{l}f(-1)\ge -4\\ \Rightarrow a-b+c\ge -4-----\left(i\right)\\ f\left(1\right)\le 0\\ \Rightarrow a+b+c\le 0\\ \Rightarrow -a-b-c\ge 0----\left(ii\right)\\ f\left(3\right)\ge 5\\ \text{ and }9a+3b+c\ge 5----\left(iii\right)\\ \text{ From (i) }+\left(\mathrm{ii}\right)\\ \Rightarrow -2b\ge -4----\left(iv\right)\\ \text{ From (ii) }+\text{ (iii) }+\text{ (iv) }\\ \Rightarrow 8a\ge 1\\ \Rightarrow a\ge 1/8\end{array}$