TITLE: If $s=a+b+c,p=ab+bc+ac, r=abc, s^2-2p+r=4$ and $a,b,c>0,$ prove that $2sp+15r^2\geq 33r$
QUESTION [0 upvotes]: If $s=a+b+c$,$p=ab+bc+ac$, $r=abc$, $s^{2}-2p+r=4$ and $a,b,c>0,$
prove that $$2sp+15r^{2}\geq 33r.$$
I tried to use some inequalities from here, but nothing works. Suggestions ?

REPLY [1 votes]: If $a,b,c \gt 0$ and $s=a+b+c$,$p=ab+bc+ac$, $r=abc$
This looks like the coefficient of some polynomial in $x$, where $ x \in \{ a,b,c \} $
$$x^3-sx^2+px-r = 0$$
From $s^2-2p+r=4$ , say $ r =4+ 2p-s^2$
$$2sp+15r^2 \geq 33r$$
$$2sp+15(4+2p-s^2)^2 - 33(4+2p-s^2) \geq 0$$
$$ 15s^4-60ps^2-87s^2+2ps+60p^2+174p+108 \geq 0$$
Noticing the number of positive constant, you can substitute back the value of $s$ and $p$, Or use that fact that $x^3-sx^2+px-4-2p+s^2= 0$, where $ x \in \{ a,b,c  \} $ to reduce the problem by removing powers of $s$ or $p$ and finish it up
$$ 15s^4+2ps+60p^2+174p+108 \geq 60ps^2+87s^2$$
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