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
 A: WLOG $a\le b\le c$ so $\frac{b+c}{2}\ge 1$
let $$f(a,b,c)= abc(a^2+b^2+c^2)$$ now $$f(a,b,c)-f(a,\frac{b+c}{2},\frac{b+c}{2})=-\frac{1}{8} a (b - c)^2 (2 a^2 + b^2 - 2 b c + c^2)\le 0$$  let $t=\frac{b+c}{2}\ge 1$ So it suffices to show  $$f(a,t,t)\le 3 \iff f(3-2t,t,t)\le 3$$ but $$f(3-2t,t,t)-3=-3 (t - 1)^2 (4 t^3 - 6 t^2 + 2 t + 1)\le 0$$ which is true  because $4t^3-6t^2+2t=2t(t-1)(2t-1)\ge 0$
Done!
A: 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!
A: 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 $(ab-bc)^2+(bc-ca)^2+(ca-ab)^2 \geq 0$.
