$a+b+c−3≥k(a−b)(b−c)(c−a)$

Question

Given that $a,b,c \ge 0$ satisfy $ab+bc+ca=3$. Find the maximum value of the real number $k$ such that the following inequality is always true:$$a+b+c-3 \ge k(a-b)(b-c)(c-a)$$

"I got $c=0,ab=3 \iff a=\frac{3}{b}$, with $a \ge b$. Then $k \le \dfrac{a^2-3a+3}{9-3a^2}$

I don't know if it's true, help me to find the maximum value of $k$ please"

Answer 1

We need to prove that $$a + b + c - 3 \ge \frac16 (a- b)(b - c)(c - a).$$

Since $(a + b + c)^2 \ge 3(ab + bc + ca) = 9$, it suffices to prove that $$(a + b + c - 3)^2 \ge \frac{1}{36}(a-b)^2(b-c)^2(c-a)^2. \tag{1}$$

We use the pqr method.

Let $p = a + b + c, q = ab + bc + ca = 6, r = abc$.

(1) is written as $$(2p^3 - 27p + 27r)^2 + 4(p + 6)(54 - p^3)(p - 3)^2 \ge 0. \tag{2}$$

If $54 - p^3 \ge 0$, clearly, (2) is true.

If $54 - p^3 < 0$, we have $2p^3 - 27p > 0$, and \begin{align*} &(2p^3 - 27p + 27r)^2 + 4(p + 6)(54 - p^3)(p - 3)^2\\ \ge{}& (2p^3 - 27p)^2 + 4(p + 6)(54 - p^3)(p - 3)^2\\ ={}&729(p-4)^2\\ \ge{}& 0. \end{align*}

We are done.

Answer 2

Yes, your attempt gives a maximal value of $k$!

But we need $a<b$, which gives: $$\min\limits_{0<a<\sqrt3}\frac{a^2-3a+3}{3(3-a^2)}=\frac{1}{6}$$ and it's enough to prove that: $$6(a+b+c-3)\geq(a-b)(b-c)(c-a).$$ Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$ and $abc=w^3$.

Thus, $v=1$ and since $$a+b+c\geq\sqrt{3(ab+ac+bc)}=3,$$ it's enough to prove that: $$36(a+b+c-3)^2\geq\prod_{cyc}(a-b)^2$$ or $$36v^4(3u-3v)^2\geq27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or $$w^6+2u(2u^2-3v^2)+(3u^2-4v^2)^2v^4\geq0,$$ which is obvious for $2u^2-3v^2\geq0.$

But for $2u^2-3v^2\leq0$ it's enough to prove that: $$u^2(2u^2-3v^2)^2-(3u^2-4v^2)^2v^4\leq0$$ or $$4(u^3-3uv^2+2v^3)(u^3-2v^3)\leq0,$$ which is true because $$u^3-2v^3\leq\left(\left(\sqrt{\frac{3}{2}}\right)^3-2\right)v^3<0$$ and by AM-GM $$u^3-3uv^2+2v^3\geq 3\sqrt[3]{u^3\cdot(v^3)^2}-3uv^2=0.$$