# $a+b+c−3≥k(a−b)(b−c)(c−a)$

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"

• Yes, your attempt gives a maximal value of $k$. But we need $a<b$, which gives: $\min\limits_{0<a<\sqrt3}\frac{a^2-3a+3}{3(3-a^2)}=\frac{1}{6}.$ Commented Dec 7, 2023 at 14:33
• thank you @MichaelRozenberg Commented Dec 7, 2023 at 14:47

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.

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

But we need $$a, which gives: $$\min\limits_{0 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 $$u^3-3uv^2+2v^3\geq 3\sqrt[3]{u^3\cdot(v^3)^2}-3uv^2=0.$$

• If I let $p=a+b+c, q=ab+bc+ca, r=abc$. I need prove: $972(p-3)^2-4(p^2-9)^3+(2p^3-27p+27r)^2 \ge 0$ help me prove that. Commented Dec 7, 2023 at 15:45
• @POQ123 It's exactly that I proved. See my solution. Commented Dec 7, 2023 at 17:48