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

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 ?

Answer 1

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 $$ 15s^4+2ps+60p^2+174p+108 \geq 60ps^2+87s^2$$