let $a,b,c>0$,and such $$a+b+c=3$$,show that $$abc(a^2+b^2+c^2)\le 3$$
My idea: since $$abc\le\left(\dfrac{a+b+c}{3}\right)^3=1$$ but $$a^2+b^2+c^2\ge \dfrac{1}{3}(a+b+c)^2=3$$ so I can't prove this inequality.Thank you
It is said that can use AM-GM inequality to solve it
Notice that $$ 3(a+b+c)abc(a^2+b^2+c^2) \leq (ab+bc+ca)^2(a^2+b^2+c^2) \leq \left(\frac{(a+b+c)^2}{3}\right)^3. $$
The first inequality follows from $a^2b^2+b^2c^2+c^2a^2 \geq abc(a+b+c)$, which is true as $a^2b^2+b^2c^2 \geq 2ab^2c$, $b^2c^2+c^2a^2 \geq 2abc^2$ and $c^2a^2+a^2b^2 \geq 2a^2bc$ by AM-GM.
The second inequality follows from AM-GM applied on $ab+bc+ca$, $ab+bc+ca$ and $a^2+b^2+c^2$.
