Prove that $3x^3-41x+48\leq 0$ for $x \in [\sqrt 3, \sqrt 6]$

Question

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!

Polynomial proof:

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.

Answer 1

We can prove this inequality by the following way.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2$.

Thus, $t\geq1$, the condition gives $3u^2-2v^2=1$ and we need to prove $$6u(3u^2-2v^2)\geq3\sqrt{(3u^2-v^2)^3}+\frac{3}{8}(9uv^2-w^3)$$ or $$48u^3-41uv^2+w^3\geq8\sqrt{(3u^2-2v^2)^3}.$$ Now, by Schur $$w^3\geq4uv^2-3u^3,$$ which says that it's enough to prove $$45u^3-37uv^2\geq8\sqrt{(3u^2-2v^2)^3}$$ or $$t(45t-37)^2\geq64(3t-2)^3$$ or $$(t-1)(297t^2+423t-512)\geq0,$$ which is obvious.

Now, you know that your polynomial inequality is true

because we solved the problem by another way (just by using Schur only).