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

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.

• Please don't tell us what to do or how to do it, or assign us work you were assigned. Instead, please read the guide to new users, on how to ask a good question and edit your post accordingly. It seems you are still asking the type of question we typically get only from newbie users, hence you need to read the post as well. Apr 23, 2020 at 12:19
• Sure. If I write the proof using polynomials then you are fine with me asking for algebraic proofs? Apr 23, 2020 at 12:21
• You must at least provide us with whatever you tried. Apr 23, 2020 at 12:22
• Is it not a contest question this time? What is the context? Apr 23, 2020 at 12:26
• Yes, please write the proof you know first. Apr 23, 2020 at 12:26

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.