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