Prove that $3x^3-41x+48\leq 0$ for $x \in [\sqrt 3, \sqrt 6]$.
This is from an inequality in one of Titu Andreescu’s inequality books. More exactly, $2(a+b+c)\geq 3+\frac38(a+b)(b+c)(c+a)$ for positive numbers with $a^2+b^2+c^2=3$. You get the inequality in the op by using the pqr method.
Please don’t use polynomial roots or continuity (I know that you can consider it a polynomial $f$ and then simply checking some values for which $f(x_0)$ is positive and negative will prove the inequality). I’m struggling to find an algebraic proof. Please help. Thank you!
Let $f(x)=3x^3-41x+48$. Then $f(-1)>0$, $f(\sqrt 3)<0$, $f(\sqrt 6)<0$ and $f(3)>0$. Thus, $f$ can’t have any roots in $[\sqrt 3, \sqrt 6] $ and we’re done.