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Given positive $a, b, c$, prove that $$\large (a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$$

As a starting point,

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

$$\iff \frac{a^2 + b^2 + c^2}{ab + bc + ca} \ge \frac{(a^3 + b^3 + c^3)(a + b + c)}{(a^2 + b^2 + c^2)^2}$$

$$\iff \frac{(a - b)^2 + (b - c)^2 + (c - a)^2}{2(ab + bc + ca)} \ge \frac{ab(a - b)^2 + bc(b - c)^2 + ca(c - a)^2}{(a^2 + b^2 + c^2)^2} \tag 1$$

$$\iff \sum_{cyc}(a - b)^2\left[\frac{1}{2(ab + bc + ca)} - \frac{ab}{(a^2 + b^2 + c^2)^2}\right] \ge 0$$

$$\iff \frac{1}{2} \cdot \sum_{cyc}\frac{[a^4 + b^4 + c^4 + 2(ca)^2 + 2(bc)^2 - 2a^2bc - 2b^2ca](a - b)^2}{(ab + bc + ca)(a^2 + b^2 + c^2)^2} \ge 0$$

$$\iff \frac{\displaystyle (a^4 + b^4 + c^4) \cdot \sum_{cyc}(a - b)^2 + 2\sum_{cyc}ca(ca - b^2)[(a - b)^2 + (b - c)^2]}{(ab + bc + ca)(a^2 + b^2 + c^2)^2} \ge 0$$

Please note for $(1)$ that $$(a^2 + b^2 + c^2) - (ab + bc + ca) = \frac{(a - b)^2 + (b - c)^2 + (c - a)^2}{2}$$ and $$(a^3 + b^3 + c^3)(a + b + c) - (a^2 + b^2 + c^2)^2 = ab(a - b)^2 + bc(b - c)^2 + ca(c - a)^2$$

Moreover, evaluating $\displaystyle \sum_{cyc}ca(ca - b^2)[(a - b)^2 + (b - c)^2]$, we have that $$\sum_{cyc}ca(ca - b^2)[(a - b)^2 + (b - c)^2] = \sum_{cyc}ca(ca - b^2)(c^2 + a^2 + 2b^2 - 2bc - 2ab)$$

$$ = \sum_{cyc}[(ca)^2 - b^2ca](c^2 + a^2) + 2bca \cdot \sum_{cyc}(ca - b^2)(b - c - a)$$

$$ = \sum_{cyc}b^2[(ab)^2 - c^2ab + (bc)^2 - a^2bc] - 2bca \cdot \sum_{cyc}b[(ab - c^2) + (bc - a^2) - (ca - b^2)]$$

I'm going to stop here. There was an attempt, one that failed tremendously.

Michael Rozenberg
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1 Answers1

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Your SOS works!

Indeed, by your work we obtain: $$(a^2+b^2+c^2)^3-(a^3+b^3+c^3)(ab+ac+bc)(a+b+c)=$$ $$=(a^2+b^2+c^2)^2(ab+ac+bc)\left(\frac{a^2+b^2+c^2}{ab+ac+bc}-\frac{(a^3+b^3+c^3)(a+b+c)}{(a^2+b^2+c^2)^2}\right)=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)^2((a^2+b^2+c^2)^2-2ab(ab+ac+bc))\geq$$ $$\geq\frac{1}{2}\sum_{cyc}(a-b)^2((a^2+b^2+c^2)(ab+ac+bc)-2ab(ab+ac+bc))=$$ $$=\frac{1}{2}(ab+ac+bc)\sum_{cyc}(a-b)^2((a-b)^2+c^2)\geq0.$$ Also, we can use $uvw$, which gives a solution immediately.

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

Thus, $a^2+b^2+c^2$, $ab+ac+bc$ and $a+b+c$ dot't depend on $w^3$ and since $$a^3+b^3+c^3=27u^3-27uv^2+3w^3,$$ it's enough to prove our inequality for a maximal value of $w^3$, which happens for equality case of two variables.

Since, our inequality is homogeneous, it's enough to assume $b=c=1$, which gives $$(a-1)^2(a^4-2a+4)\geq0$$ and we are done!

Michael Rozenberg
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  • How homogeneity is sufficient to assume $b=c=1$? – Quang Hoang Jul 17 '19 at 16:00
  • @Quang Hoang After replacing $a$ on $ka$, $b$ on $kb$ and $c$ on $kc$ for all $k>0$ our inequality is not changed. Thus, we can choose a value of $k$ such that $kb=kc=1.$ Remember: We work with the equality case of two variables. – Michael Rozenberg Jul 17 '19 at 16:10
  • I don't see where you show $b=c$. You did mention something about maximal value of $w^3$, but that happens when $a=b=c$? – Quang Hoang Jul 17 '19 at 16:15
  • In the case $a=b=c$ the inequality is obvious. – Michael Rozenberg Jul 17 '19 at 16:15
  • Of course, but how does that relate to the statement in discussion, i.e. $b=c$? – Quang Hoang Jul 17 '19 at 16:17
  • @Quang Hoang Do you want that I'll prove that $w^3$ gets a maximal value for equality case of two variables? I am ready to show. – Michael Rozenberg Jul 17 '19 at 16:18
  • Actually, what do you mean by '$w^3$ gets a maximal value for equality case of two variables'. I only see that $w^3$ is maximal when $a=b=c$. – Quang Hoang Jul 17 '19 at 16:23
  • @Quang Hoang We need to prove that $(9u^2-6v^2)^3\geq(27u^3-27uv^2+3w^3)3v^2\cdot3u.$ Now let $u$ and $v^2$ be constants and $w^3$ is changed. We see that it's enough to prove the last inequality for a maximal value of $w^3$. – Michael Rozenberg Jul 17 '19 at 16:24
  • right, and how that relates to $b=c$? – Quang Hoang Jul 17 '19 at 16:26
  • @Quang Hoang An equality case of two variables it's or $a=b$ or $a=c$ or $b=c$. Our inequality is symmetric. It's enough to assume $b=c$ – Michael Rozenberg Jul 17 '19 at 16:29
  • And, your whole statement might be incorrect. Consider an example: minimize $u-v + w$ where $w=(u-v+2)^2 +a^2$. Your argument says $w$ is minimal when $u-v=-2$ and $a=0$. So the answer is $-2$. Is it a valid argument? – Quang Hoang Jul 17 '19 at 16:30
  • again WHY $w^3$ is maximal when there are two equal variables? It MUST be three equal variable for $w^3$ to be maximal. – Quang Hoang Jul 17 '19 at 16:32
  • @Quang Hoang There are very many wrong statements. I think, it's better to say about our problem. We wrote $(a^2+b^2+c^2)^3-(a^3+b^3+c^3)(ab+ac+bc)(a+b+c)$ as a polynomial of new not depend variables $u$, $v^2$ and $w^3$. For all symmetric polynomial with three variables we can do it. Now, about your last question. If we'll prove that our inequality is true for $b=c$, so it must be true for $a=b=c$. – Michael Rozenberg Jul 17 '19 at 16:34
  • Actually, I started to understand your point. Are you trying to say that, given $u$ and $v$ (fixed), then $w$ is maximal when two of the three variables are equal? – Quang Hoang Jul 17 '19 at 16:38
  • @Quang Hoang Yes, of course! It's better to say about $u$, $v^2$ and $w^3$ and variables, for which two of them need be equal are $a$, $b$ and $c$. – Michael Rozenberg Jul 17 '19 at 16:40