I need to prove something like that:
For $a,b,c>0$ prove: $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$.
I know that $3abc≤(a^3+b^3+c^3)$, but then I derived $3(ab+bc+ca) ≤ (a+b+c)^2$, I can't move on.
Can anyone help me?
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Sign up to join this communityI need to prove something like that:
For $a,b,c>0$ prove: $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$.
I know that $3abc≤(a^3+b^3+c^3)$, but then I derived $3(ab+bc+ca) ≤ (a+b+c)^2$, I can't move on.
Can anyone help me?
Hint: Using CS inequality,
$$(a^3+b^3+c^3)\left(\frac1a+\frac1b+\frac1c\right)\geqslant (a+b+c)^2$$
Hint:
$$-(a+b)(a+c)(b+c)(a^2-ab-ac+b^2-bc+c^2)=abc(a+b+c)^2-(a^3+b^3+c^3)(ab+bc+ca)$$