If $a^3+b^3+c^3=3$ so $2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$

Question

I was looking at this question, and I derived this inequality from that

Let $a$, $b$ and $c$ be positive numbers such that $a^3+b^3+c^3=3.$ Prove that:

$$2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$$

I couldn't solve Rozenberg's inequality but I assumed it's true. So if $a^3+b^3+c^3=3$, then

$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}\tag {1}$$

and also by using this, if $b^3+a^3+c^3=3$, then

$$\frac{b^3}{b+a}+\frac{a^3}{a+c}+\frac{c^3}{c+b}\geq\frac{3}{2}\tag {2}$$

using $(1)$ and $(2)$, if $a^3+b^3+c^3=3$, then

$$\frac{a^3+b^3}{a+b}+\frac{b^3+c^3}{b+c}+\frac{c^3+a^3}{c+a}\geq3$$

$$\Leftrightarrow \sum\limits_{cyc}(a^2-ab+b^2)\geq 3$$

$$\Leftrightarrow 2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$$

If I'm not making any silly mistake, this inequality should be true. I couldn't solve yet, but I wanted to share with you.

$\mathbf {EDIT:}$ To be clear, I think that I derived an inequality weaker than Rozenberg's inequality by using his inequality, and I presented it for those of you who might interest trying to solve inequalities by ownselves

Answer 1

We need to prove that $$\left(\sum_{cyc}(2a^2-ab)\right)^3\geq3(a^3+b^3+c^3)^2$$ or $$\sum_{cyc}(5a^6-12a^5b-12a^5c+30a^4b^2+30a^4c^2-31a^3b^3-15a^3b^2c-15a^3c^2b+20a^2b^2c^2)\geq0$$ or $$\sum_{cyc}(10a^6-24a^5b-24a^5c+60a^4b^2+60a^4c^2-62a^3b^3-30a^3b^2c-30a^3c^2b+40a^2b^2c^2)\geq0$$ or $$5\sum_{cyc}(a^6-a^5b-ab^5+b^6)-19\sum_{cyc}(a^5b-a^4b^2-a^2b^4+ab^5)+$$ $$+41\sum_{cyc}(a^4b^2+a^2b^4-2a^3b^3)+10\sum_{cyc}(2a^3b^3-a^3b^2c-a^2c^2b)-$$ $$-20abc\sum_{cyc}(a^2b+a^2c-2abc)\geq0$$ or $$\sum_{cyc}(a-b)^2(5(a^4+a^3b+a^2b^2+ab^3)-19ab(a^2+ab+b^2)+41a^2b^2+20c^3(a+b)-20abc^2)\geq0$$ or $$\sum_{cyc}(a-b)^2(20(a+b)c^3-20abc^2+5a^4-14a^3b+21a^2b^2-14ab^3+5b^4)\geq0.$$ Now, Let $c=x\sqrt{ab}$.

Hence, since $$5a^4-14a^3b+21a^2b^2-14ab^3+5b^4-3a^2b^2=(a-b)^2(5a^2-4ab+5b^2)\geq0,$$ by AM-GM we obtain: $$20(a+b)c^3-20abc^2+5a^4-14a^3b+21a^2b^2-14ab^3+5b^4\geq$$ $$\geq40\sqrt{ab}c^3-20abc^2+3a^2b^2=a^2b^2(40x^3-20x^2+3)=$$ $$=a^2b^2(20x^3+20x^3+3-20x^2)\geq a^2b^2\left(3\sqrt[3]{\left(20x^3\right)^2\cdot3}-20x^2\right)=$$ $$=20a^2b^2x^2\left(3\sqrt[3]{1200}-20\right)>0$$ and we are done!

Also we can use the $uvw$ method.

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

Hence, $\sum\limits_{cyc}(2a^2-ab)$ does not depend on $w^3$ and

$a^3+b^3+c^3=27u^3-27uv^2+3w^3$ is an increasing function of $w^3$.

Thus, it's enough to prove our inequality for a maximal value of $w^3$,

which happens for equality case of two variables

and $b=c=1$ in the homogeneous form gives $$(2(a^2+2)-2a-1)^3\geq3(a^3+2)^2$$ or $$(a-1)^2(5a^4-14a^3+27a^2-24a+15)\geq0,$$ which is obvious and we are done again.

About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791

Answer 2

Rozenberg's inequality is solved, find the solution at this link :

If $a^3+b^3+c^3=3$ so $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}$

Answer 3

Alternative proof: the pqr method

Let $p = a + b + c$ and $q = ab + bc + ca$ and $r = abc$. We will use $p^2 \ge 3q$ and $q^2 \ge 3pr$ which are well-known. See the remarks at the end.

Using $a^3 + b^3 + c^3 = p^3 - 3pq + 3r$, the condition is written as $$p^3 - 3pq + 3r = 3. \tag{1}$$

Using (1) and $q^2 \ge 3pr$, we have $$p^3 - 3pq + 3\cdot \frac{q^2}{3p} \ge 3$$ or $$\frac{1}{p}\left(\frac32p^2 - q\right)^2 \ge \frac54 p^3 + 3$$ which results in (using $p^2 \ge 3q$) $$\frac32p^2 - q \ge \frac12\sqrt{5p^4 + 12p}$$ or $$q \le \frac{3}{2}p^2 - \frac12\sqrt{5p^4 + 12p}. \tag{2}$$

Using (2), we have \begin{align*} &2(a^2 + b^2 + c^2) - (ab + bc + ca) - 3\\ =\,& 2(p^2 - 2q) - q - 3\\ =\,& 2p^2 - 5q - 3\\ \ge\,& 2p^2 - 5\cdot \left(\frac{3}{2}p^2 - \frac12\sqrt{5p^4 + 12p}\right) - 3 \\ =\,& \frac52\sqrt{5p^4 + 12p} - \frac{11}{2}p^2 - 3\\ \ge\,& 0 \end{align*} where we have used $$\left(\frac52\sqrt{5p^4 + 12p} \right)^2 - \left( \frac{11}{2}p^2 + 3\right)^2 = (3 - p)(-p^3 - 3p^2 + 24p - 3) \ge 0.$$ (Note: It is easy to prove that $1 \le p \le 3$. Also, $-p^3 - 3p^2 + 24p - 3 \ge 0$ for all $1 \le p \le 3$. See the remarks at the end.)

We are done.

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Remarks:

1. $p^2 \ge 3q$ is just $a^2 + b^2 + c^2 \ge ab + bc + ca$.

2. $q^2 \ge 3pr$ is just $(ab)^2 + (bc)^2 + (ca)^2 \ge ab \cdot bc + bc \cdot ca + ca \cdot ab$.

3. We have $p \le 3$ which follows from the power mean inequality $\sqrt[3]{\frac{a^3 + b^3 + c^3}{3}} \ge \frac{a + b + c}{3}$.

4. We have $p \ge 1$. Indeed, if $p < 1$, then $a, b, c \le 1$ and $1 > a + b + c \ge a^3 + b^3 + c^3$ which contradicts the condition.