How prove this $(abc)^4+abc(a^3c^2+b^3a^2+c^3b^2)\le 4$
It is possible to solve this by $pqr$ method. (That is, write everything in terms of $p = a+b+c$, $q=ab+bc+ca$, $r = abc$) together with $uvw$ method. This is an ugly method, but it works. The approach is sketched here.
- Although the inequality is not symmetric, one can write $$\begin{array} &&2(a^3c^2 + b^3a^2 + c^3b^2) \\ &= a^3(b^2+c^2) + b^3(a^2+c^2) + c^3(a^2+b^2) + (a^3c^2+b^3a^2 + c^3b^2 - a^3b^2 - b^3c^2 - c^3a^2) \\ &= a^3(b^2+c^2) + b^3(a^2+c^2) + c^3(a^2+b^2) + (b-a)(c-a)(c-b)(ab+bc+ca) \end{array}$$
This allows one to rewrite the inequality as $$2(abc)^4 + abc(\sum_{sym} a^3b^2) + abc(ab+bc+ca)\sqrt{(a-b)^2(b-c)^2(c-a)^2} \leq 8$$
- I used Sage to express the expression in terms of $p,q,r$, which hopefully someone can verify:
$$2r^4 + r(pq^2 - 2p^2r - qr) + qr \sqrt{\frac{4(p^2-3q)^3 - (2p^3-9pq+27r)^2}{27}} \leq 8$$
The main problem here, would be to efficiently remove the square root. For this, we use a lemma by Vasile Cirtoaje and Vo Quoc Bao Can, lemma 2.1 in here. In the lemma we take $$\alpha = \frac{1}{\sqrt{27}}, a = 2(9-3q)^{3/2}, b = 54-27q+27r, \beta = \frac{14+q}{27q}$$ (The choice of $\alpha,a,b$ comes from the shape of the square root, while the choice of $\beta$ comes from cancelling the "linear term" of $r$ in $pq-2p^2r-qr = 3q - (18+q)r$; remember that $p = 3$ in this question) Therefore LHS is bounded above by $$2r^4 + 3rq^2 + (27q\frac{14+q}{27q} - (18+q))r^2 + r(q\frac{14+q}{27q} (54-27q) + 2(9-3q)^{3/2} \sqrt{\frac{q^2}{27} + \left(\frac{14+q}{27}\right)^2}) ...(*)$$ and it suffices to show that $(*) \leq 8$.
For this, we use the uvw method. Fix $r$. By the help of a computer, one can check that $(*)$ is convex in $q$. Thus to maximize $(*)$, $q$ should take maximum/minimum of its allowable range, which would force two of $a,b,c$ to be equal according to $uvw$ method.
Assume WLOG that $a=c$. Then $2a+b = 3$, and $a^2b = r$. Solving these gives $$b = 3-2a \, \text{ and } \, r = a^2(3-2a) \, \text{ and } \, q = -3a^2+6a$$ The only constraint we have on $a$ are $a \in [0,3/2]$ (coming from $2a+b=3$ and $a,b \ge 0$), and $a^2(3-2a) = r \leq 1$ (since $p = 3$). The latter inequality is equivalent to $a \ge -1/2$, which means that the only constraint we have on $a$ is $a \in [0,3/2]$.
Therefore we are reduced to showing that for $a \in [0,3/2]$, the substituted expression $(**) \leq 8$, where $(**)$ comes from substituting $q,r$ into $(*)$ in terms of $a$. I find that $(**)$ is the following, and please verify this if (quite unlikely) you are reading up to this line. $$2 (a^2 (3-2 a))^4+3a^2(3-2a)(-3a^2+6a)^2 - 4(a^2(3-2a))^2 + a^2(3-2a)(14 - 3a^2 + 6a)(2 - (-3a^2+6a))+2a^2 (3-2 a) (9-3 (-3 a^2+6 a))^{3/2} \sqrt{\frac{(-3a^2+6a)^2}{27}+\frac{(14-3 a^2+6 a)^2}{27^2}}..(**)$$ This is a single variable calculus problem that is ugly but should be tractable. I trust WolframAlpha to do the job for me, which does give the maximum is 8.
EDit: 5th version: After read "uvw method", I found the trick is:$abc(a^3c^2+b^3a^2+c^3b^2)\leq 3$, but I have assumption that $a \ge b \ge c$,and all my explore is based on it but it is wrong. thanks Sanchez points out my mistake which force me to recheck everything and I get a real result $a=1.3175,b=0.5856,c=1.0969$ that cause $abc(a^3c^2+b^3a^2+c^3b^2)=3.00667>3$ ,Indeed, the $abc(a^3c^2+b^3a^2+c^3b^2)\leq 3$ will be true when $(a-b)(b-c)(c-a) \leq 0 $, if $(a-b)(b-c)(c-a) \ge 0 $, then $abc(a^3b^2+b^3c^2+c^3a^2)\leq 3$ will be true.that is a real trick i found. To proof it ,just use Sanchez's work:
$2abc(a^3c^2+b^3a^2+c^3b^2)=abc(\sum_{sym} a^3b^2) + abc(ab+bc+ca)(a-b)(b-c)(c-a) \leq 6$, I will proof $abc(\sum_{sym} a^3b^2) \leq 6$ next. then the result is clear. if we swap $b,c$ , we have another result at once.
Here I still use Lagrange multiplier to proof $abc(a^3c^2+b^3a^2+c^3b^2+a^3b^2+a^2c^3+b^3c^2) \leq 6$.
let $f=abc(a^3c^2+b^3a^2+c^3b^2+a^3b^2+a^2c^3+b^3c^2)- 6$, $g=a+b+c-3$, $F=f-\lambda g ,p=a^3c^2+b^3a^2+c^3b^2+a^3b^2+a^2c^3+b^3c^2,$
$\dfrac{\partial F}{\partial a}=bcp+abc[3a^2(b^2+c^2)+2a(b^3+c^3)]= \lambda $......<1>
$\dfrac{\partial F}{\partial b}=acp+abc[3b^2(a^2+c^2)+2b(a^3+c^3)]= \lambda $......<2>
$\dfrac{\partial F}{\partial c}=abp+abc[3c^2(a^2+b^2)+2c(a^3+b^3)]= \lambda $......<3>
$\dfrac{\partial F}{\partial \lambda}=a+b+c-3=0$......<4>
<1> - <2> : $pc(b-a)+abc[3c^2(a^2-b^2)+2ab(b^2-a^2)+2c^3(a-b)]=0$
that is:
$c(a-b)[3abc^2(a+b)+2c^3ab-(3a^2b^2(a+b)+a^2c^2(a+c)+b^2c^2(b+c))]=0$......<5>
so we have $c=0$ or $a=b$ or
$3abc^2(a+b)+2c^3ab=3a^2b^2(a+b)+c^2(a^3+b^3)+c^3(a^2+b^2)$......<6>,
we look at <6> carefully,
$c^3(a^2+b^2) \ge 2c^3ab $, $3a^2b^2(a+b) \ge 3abc^2(a+b)$,so RHS $> $ LHS, so we can ignore<6>.
the $c=0$ is the max value of $a^3c^2+b^3a^2+c^3b^2+a^3b^2+a^2c^3+b^3c^2$, that is why $a^3c^2+b^3a^2+c^3b^2<3$ is not true. we can ignore here as it will lead f=-6. so we have a=b,
In case $a=b $, let <1>-<3>, we have:
$b(a-c)[p+bc(2ac(a+c)-3b^2(a+c)-2b^3)]=0$......<7>,
we have $a=c$, or $[p+bc(2ac(a+c)-3b^2(a+c)-2b^3)]=0$ which is:
$3a^2c^2(a+c)+b^3(a^2+c^2)+b^2(a^3+c^3)=3ab^3c(a+c)+2ab^3c$......<8>,
put $a=b, 2b+c=3$, we simplify<8> to:
$10 b^3-84 b^2+153 b-81 = 0 $......<9>,
$q(b)=10 b^3-84 b^2+153 b-81,q'(b)=30b^2-168b+153=0,$......<10>
get two roots: $b_1 = \dfrac{1}{10}(28 - \sqrt{274}), b_2 = \dfrac{1}{10}(28 + \sqrt{274}>4)$,
put $b=0$ and $b=2$ into<10>,$q'(0)>0,q'(2)<0,$ so $b_1$ is max point between $(0,2)$, put $b_1$ into $q(b_1) =-.9<0$ ,so there is no solution for $0<b \leq 1.5$ so the only solution is $a=c$ for <7>.
last step, we verify , when $a=b=c=1,f=0,$ we already have $-6$ before,so $f_{max}=0$. QED.
I copy a special method to prove $abc+abc(a^3b^2+b^2c^3+c^3a^2) \leq 4$ which is equal to current one. $abc+abc(a^3b^2+b^2c^3+c^3a^2) \leq 4$$ \Leftrightarrow4(a+b+c)^8-6561abc(a^3c^2+b^3a^2+c^3b^2)-27abc(a+b+c)^5\geq0 $
$ a=\min\{a,b,c\} , b=a+u ,c=a+v $
$ 4(a+b+c)^8-6561abc(a^3c^2+b^3a^2+c^3b^2)-27abc(a+b+c)^5= $$ =4374(u^2-uv+v^2)a^6+243(49u^3+21u^2v-60uv^2+49v^3)a^5+ 81(164u^4+242u^3v-168u^2v^2-163uv^3+164v^4)a^4+ 27(208u^5+707u^4v+352u^3v^2-1106u^2v^3+221uv^4+208v^5)a^3+9(109u^6+609u^5v+1455u^4v^2-1006u^3v^3-732u^2v^4+609uv^5+109v^6)a^2+ +3(32u^7+215u^6v+627u^5v^2+1030u^4v^3-1157u^3v^4+627u^2v^5+215uv^6+32v^7)a+4(u+v)^8\geq0 $
......<1>
let $K_{i}$means $K_{i}a^i, i=1 to 6$, If $K_{i} \ge 0$, then <1> will be true.
$K_6=u^2-uv+v^2 \ge 0$ is trivial . $K_5=49u^3+21u^2v-60uv^2+49v^3$, $21u^2v+49v^3 \ge 2\sqrt{21*49}uv^2>2*32uv^2>60uv^2$ $\to K_5 \ge 0$ $K_4=164u^4+242u^3v-168u^2v^2-163uv^3+164v^4=164u^4+78u^3v-4u^2v^2+uv^3+164v(u^3-u^2v-uv^2+v^3)$, $(u-v)(u^2-v^2) \ge 0 \to u^3-u^2v-uv^2+v^3 \ge 0, 78u^3v+uv^3 \ge 2\sqrt{78*1}u^2v^2 > 4u^2v^2 \to K_4 \ge 0$ $K_3=208u^5+707u^4v+352u^3v^2-1106u^2v^3+221uv^4+208v^5$,$352u^3v^2+221uv^4 \ge 2\sqrt{352*221}u^2v^3>557u^2v^3,707u^4v+208v^5 \ge 2\sqrt{707*208}u^2v^3>766u^2v^3,557+766=1323>1106 \to K_3 \ge 0$ $K_2=109u^6+609u^5v+1455u^4v^2-1006u^3v^3-732u^2v^4+609uv^5+109v^6$,$609u^5v+609uv^5 \ge 1218u^3v^3>1006u^3v^3,1455u^4v^2+109v^6 \ge 2\sqrt{1455*109}u^2v^4>796u^2v^4>732u^2v^4 \to K_2 \ge 0$ $K_1=32u^7+215u^6v+627u^5v^2+1030u^4v^3-1157u^3v^4+627u^2v^5+215uv^6+32v^7,1030u^4v^3+627u^2v^5 \ge 2\sqrt{1030*627}u^3v^4 >1607u^3v^4 >1157u^3v^4 \to K_1 \ge 0 $
QED.
the original proof is here