Find max $P=a^3b^2+b^3c^2+c^3a^2+7(ab+bc+ca).$

Question

Let $a,b,c\ge 0: a+b+c=3.$ Find the maximum of $$P=a^3b^2+b^3c^2+c^3a^2+7(ab+bc+ca)\le 24.$$ I set $a=b=c=1$ and got $P=24.$

My approach is proving $$a^3b^2+b^3c^2+c^3a^2+7(ab+bc+ca)\le 24$$ Firstly, I tried to give identity $$2(a^3b^2+b^3c^2+c^3a^2)=\sum_{cyc}a^2b^2(a+b)+\sum_{cyc}a^3(b^2-c^2)=\sum_{cyc}a^2b^2(a+b)+(ab+bc+ca)(a-b)(b-c)(a-c)=3(ab+bc+ca)^2-(18+ab+bc+ca)abc+(ab+bc+ca)(a-b)(b-c)(a-c)$$ But I don't know how to continue it. It is hard with the yield $(a-b)(b-c)(a-c)$

Hope to see some useful help here. Thank you

Answer 1

Proof.

We will prove that $24$ is the maximal value and attained at $a=b=c=1.$

I give a proof based on your idea to prove$$a^3b^2+b^3c^2+c^3a^2+7 (ab+bc+ca)\le 24.$$ Now, let the notation $a+b+c=p=3; ab+bc+ca=q; abc=r.$

By your thought progress, we'll prove $$3q^2-(18+q)r+q(a-b)(b-c)(a-c)+14q\le 48.\tag{*}$$ Notice that we already have the following identity \begin{align*} (a-b)^2(b-c)^2(a-c)^2&=p^2q^2-4q^3+2pr(9q-2p^2)-27r^2\\&=9q^2-4q^3+54r(q-2)-27r^2\\&=4(3-q)^3-27(q-2-r)^2\\&\ge 0. \end{align*} Also, for all $a,b,c\in R$ $$(a-b)(b-c)(a-c)\le \sqrt{(a-b)^2(b-c)^2(a-c)^2}.$$Thus, it's enough to prove the $(*)$ according $$3q^2-(18+q)r+q\sqrt{4(3-q)^3-27(q-2-r)^2}+14q-48\le0.\tag{**}$$ We split $(**)$ into two cases

• $0\le q\le \dfrac{24}{11}.$

Apply the well-known result: $a^2b+b^2c+c^2a+abc\le \dfrac{4}{27}(a+b+c)^3,$ which was discussed here, we obtain$$a^3b^2+b^3c^2+c^3a^2\le (ab+bc+ca)(a^2b+b^2c+c^2a)\le 4(ab+bc+ca).$$It means $$a^3b^2+b^3c^2+c^3a^2+7 (ab+bc+ca)\le 11 (ab+bc+ca)\le 24.$$

• $\dfrac{24}{11}\le q\le 3.$

Let $0\le t=3-q\le \dfrac{9}{11}$ and $s=q-2-r.$ It implies $q=3-t; r=1-s-t.$

The $(**)$ turns out $$3(3-t)^2-(21-t)(1-s-t)+14(3-t)+(3-t)\sqrt{4t^3-27s^2}-48\le 0,$$or $$10t-2t^2-s(21-t)\ge (3-t)\sqrt{4t^3-27s^2}\ge 0.$$It is equivalent to$$[10t-2t^2-s(21-t)]^2- (3-t)^2(4t^3-27s^2)\ge 0,$$or $$[(21-t)^2+27(3-t)^2]s^2-2(10t-2t^2)(21-t)s+(10t-2t^2)^2-4t^3(3-t)^2\ge 0. \tag{***}$$ Can you go futher ?

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We see that $(***)$ is a quadratic of polynomial according $s.$

Thus, it's enough to prove $$\Delta^{'}_{s}=(10t-2t^2)^2(21-t)^2-[(10t-2t^2)^2-4t^3(3-t)^2][(21-t)^2+27(3-t)^2]\le 0,$$ or $$27(3-t)^2(10t-2t^2)^2- 4t^3(3-t)^2[(21-t)^2+27(3-t)^2]\ge0$$ $$\iff 27(5-t)^2-27t(3-t)^2-t(21-t)^2\ge 0,$$ which is rewritten as $$f(t)=-28t^3+231t^2-954t+675\ge 0, t\in\left[\frac{9}{11};3\right]$$where $$f'(t)=-2(42t^2-231t+477)<0, \forall t\in R.$$It means that $$f(t)>f\left(\frac{9}{11}\right)>0.$$ The proof is done.

Answer 2

It is equivalent to $$ \frac{1}{124302682296}\sum a b^{2} \cdot \left(86733660 \left(- a c + b^{2}\right)^{2} + \left(96443 a b + 22997 a c - 119440 b^{2}\right)^{2} + 96443 \left(1326 a^{2} - 1285 a b + 377 a c - 418 b^{2}\right)^{2}\right) + \frac{1}{12421674}\sum a c^{2} \cdot \left(36051 a^{2} \left(a - b\right)^{2} + a^{2} \left(- 473 a + 54 b + 419 c\right)^{2} + 838 \left(- 19 a^{2} + 17 a b - 59 a c + 61 c^{2}\right)^{2}\right) + \frac{1}{2033424}\prod a \sum \left(- 2092 a^{2} + 564 a b - 687 a c + 123 b c + 2092 c^{2}\right)^{2} + \frac{8}{81}\sum a \left(a^{3} - a b^{2} - a b c + a c^{2} - 2 b^{3} + b^{2} c + 3 b c^{2} - 2 c^{3}\right)^{2}+ \frac{8279}{25104}\prod a \sum a^{2} \left(b - c\right)^{2}\ge 0 $$

Answer 3

For the continue you need to use the $uvw$'s technique:

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

Thus, we need to prove that: $$a^3b^2+b^3c^2+c^3a^2+21u^3v^2\leq24u^5$$ or $$48u^5-42u^3v^2-\sum_{cyc}(a^3b^2+a^3c^2)\geq\sum_{cyc}(a^3b^2-a^3c^2)$$ or $$48u^5-42u^3v^2-(27uv^4-18u^2w^3-3v^2w^3)\geq(ab+ac+bc)(a-b)(a-c)(b-c)$$ or $$16u^5-14u^3v^2-9uv^4+(6u^2+v^2)w^3\geq v^2(a-b)(a-c)(b-c).$$

We'll prove that $$16u^5-14u^3v^2-9uv^4+(6u^2+v^2)w^3\geq0.$$ Indeed, it's a linear inequality of $w^3$ and by $uvw$ it's enough to check this inequality in the following cases.

1. $w^3=0$.

Let $c=0$ and $b=1$.

Thus, we need to prove that: $$\frac{16(a+1)^5}{243}-\frac{14(a+1)^3a}{81}-\frac{(a+1)a^2}{3}\geq0$$ or $$(a+1)(16a^4+22a^3-69a^2+22a+16)\geq0,$$ which is true by AM-GM: $$16a^4+22a^3-69a^2+22a+16\geq$$ $$\geq(2\cdot16+2\cdot22-69)a^2=7a^2\geq0.$$

2. two variables are equal.

Let $b=c=1$.

Here we obtain: $$\frac{16(a+2)^5}{243}-\frac{14(a+2)^3(2a+1)}{81}-\frac{(a+2)(2a+1)^2}{3}+\left(\frac{2(a+2)^2}{3}+\frac{2a+1}{3}\right)a\geq0$$ or $$(a-1)^2(8a^3+54a^2+66a+7)\geq0.$$ Thus, it's enough to prove $$(16u^5-14u^3v^2-9uv^4+(6u^2+v^2)w^3)^2\geq27v^4(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or $$(9u^4+3u^2v^2+7v^4)w^6+u(48u^6-34u^4v^2-7u^2v^4-45v^6)w^3+$$ $$+64u^{10}-112u^8v^2-23u^6v^4+63u^4v^6+27v^{10}\geq0.$$ Now, let $u^2=tv^2$.

Thus, $t\geq0$ and since by AM-GM $$64u^{10}-112u^8v^2-23u^6v^4+63u^4v^6+27v^{10}=v^{10}(64t^5-112t^4-23t^3+63t^2+27)=$$ $$=v^{10}(8t-7)(8t^4-7t^3-9t^2+27)=$$ $$=v^{10}(8t-7)((5t^4-9t^2+5)+3t^4-7t^3+22)\geq$$ $$\geq v^{10}(8t-7)\left(4\sqrt[4]{1^3\cdot22}-7\right)t^3\geq0,$$ our inequality is true for $$48t^3-34t^2-7t-45\geq0.$$ But for $48t^3-34t^2-7t-45\leq0$ and $t\geq1$ it remains to prove that: $$t(48t^3-34t^2-7t-45)^2-4(9t^2+3t+7)(64t^5-112t^4-23t^3+63t^2+27)\leq0$$ or $$(t-1)^2(28-7t+36t^2-32t^3)\geq0,$$ which is true for $48t^3-34t^2-7t-45\leq0$ and $t\geq1$.