<s>[INST] 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
 [/INST] Assume $a=\min\{a,b,c\} \rightarrow 0\le a \le 1.$
By AM-GM, we have $$abc\left(a^2+b^2+c^2\right) =\dfrac{1}{3} a\cdot
3bc\cdot\left(a^2+b^2+c^2\right)$$
$$\le \dfrac{1}{12} a \left[a^2+bc+\left(b+c\right)^2\right]^2\le \dfrac{1}{12} a \left[a^2+\dfrac{\left(3-a\right)^2}{4}+\left(3-a\right)^2\right]^2\le 3,$$
so it suffices to prove
$$\left( 9\,{a}^{2}-33\,a+64 \right)  \left(1-a \right) ^{3}\ge 0,$$
which is obvious since $0\le a\le 1.$
Done!
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