How prove this inequality $a^3b+b^3c+c^3a+a^3b^3+b^3c^3+c^3a^3\le 6$ let $a,b,c>0$, and such
$$a^2+b^2+c^2=3$$
show that
$$a^3b+b^3c+c^3a+a^3b^3+b^3c^3+c^3a^3\le 6\tag{2}$$
I know this famous inequality( creat by valsie)
$$(a^2+b^2+c^2)^2\ge 3(a^3b+b^3c+c^3a)$$  this inequality if and only if
$$a=b=c, or,a:b:c=\sin^2{\dfrac{\pi}{7}}:\sin^2{\dfrac{2\pi}{7}}:\sin^2{\dfrac{3\pi}{7}}$$
can see:
Vasc inequality solution
so
$$(a^3b+b^3c+c^3a)\le \dfrac{1}{3}(a^2+b^2+c^2)^2=3$$
so we only prove
$$a^3b^3+b^3c^3+a^3c^3\le 3 \tag{1}$$
but this (1) inequality is not true.such as
$$a=\dfrac{5}{\sqrt{17}},b=\dfrac{5}{\sqrt{17}},c=\dfrac{1}{\sqrt{17}}$$
but 
$$a^3b^3+c^3a^3+b^3c^3\approx 3.23$$
see :inequality 
so How prove this inequality (2),this (2) inequality is true, because I have use Maple test it.
Ps: when I join in the stack exchange,it is always appear
"Mathematics Stack Exchange requires external JavaScript from another domain, which is blocked or failed to load",That's why? then I can't Can't comment
Hello,chenbai,why it is clear $A,B,C,D,E,F\ge 0$, and this is not me download it. Thank you

both side multiply $a^2+b^2+c^2=3$, we have:
$(a^2+b^2+c^2)(a^3b+b^3c+c^3a)+3(a^3b^3+b^3c^3+c^3a^3)\le \dfrac{2}{3}(a^2+b^2+c^2)^3 \iff 2(a^2+b^2+c^2)^3-3((a^2+b^2+c^2)(a^3b+b^3c+c^3a)+3(a^3b^3+b^3c^3+c^3a^3)) \ge 0 $
let $x= $min{$a,b,c$}, the other two are $x+u,x+v$, we will always have 
$Ax^4+Bx^3+Cx^2+Dx+F \ge 0 $
$A=12u^2-12uv+12v^2 \\ B=31u^3-36u^2v-9uv^2+31v^3 \\C=27u^4-21u^3v-33u^2v^2+6uv^3+27v^4 \\D=9u^5-3u^4v-12u^3v^2-15u^2v^3+12uv^4+9v^4 \\ F=2v^6+6u^2v^4-12u^3v^3+6u^4v^2-3u^5v+2u^6$
$u,v$ may swap each other when $x=a$ or $b$ or $c$
it is not difficult to prove $A,B,C,D,F \ge 0 $ and when and only when $u=v=0$ to get $LHS=0$
QED
