given $a+b+c=3$ prove that $abc(a^2+b^2+c^2) \leq 3$ I am preparing for inmo and I came accross this problem while solving a worksheet, but couldn't solve it, pl help me...
Problem-

Prove that if a,b,c are non negative real numbers such that a+b+c=3, then $abc(a^2+b^2+c^2)\leq3$ ... (0)

Developments-
Firstly applying am-gm on $a+b+c$ to get $abc\leq1$
Then applying quadratic mean arthematic mean to get
$a^2+b^2+c^2 \ge 3$
Then writing the expansion of $(a+b+c)^2$ we get
$a^2+b^2+c^2 = 9-2(ab+bc+ca)$ ...(I)
Thn applying am-hm on $ab+bc+ca$ to get
$ab+bc+ca \ge 3abc$
Then substituting in (I) we get
$a^2+b^2+c^2 \leq 9-6(abc)$
Then we substituting in (0)
$LHS \leq abc(9-6abc)$
Whose max value is $27/8$ but is not less than three, so I am stuck for this point onwards

I will use a well known inequality for my solution -
$a^2b^2+b^2c^2+c^2a^2 \geq abc(a+b+c)$
$\implies (ab+bc+ca)^2 \geq 3abc(a+b+c) = 9abc$
Now using AM-GM,
$a^2+b^2+c^2+2 (ab + bc + ca) \geq 3 [(a^2+b^2+c^2)(ab+bc+ca)^2]^{1/3}$
$9 \geq 3[(a^2+b^2+c^2)(ab+bc+ca)^2]^{1/3}$
$3^3 \geq 9 abc(a^2+b^2+c^2)$
$3 \geq abc(a^2+b^2+c^2)$

On the inequality I used to answer comes from $(ab-bc)^2+(bc-ca)^2+(ca-ab)^2 \geq 0$.
