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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 ?

Especially Lime
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Josie McLaren
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1 Answers1

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

Aderinsola Joshua
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  • You mean substitute s and p with the their components a,b,c ? This looks bads. I tried too many substitution, but calculations look bad, so I gave up. Are you sure that this is work? If I will start to do this ugly calculations, I want to be sure that this is a good method. Or you can show me a complete proof ? – Josie McLaren Mar 17 '21 at 15:20
  • The inequality already looks true, since $a,b,c \gt 0$ , then $ p \gt s $ – Aderinsola Joshua Mar 17 '21 at 15:29
  • That is not true. Is $s>=p$. For a proof, see the above link. – Josie McLaren Mar 17 '21 at 15:36
  • If you look at the inequality, if $s \ge p$, look at the powers and size, for positive $p,s$ and compare the right hand side with the left hand side, you'll know if it's true or not – Aderinsola Joshua Mar 17 '21 at 18:37
  • Comparing them is the point, but how? I don't think solving an inequality is as simple as you say.The inequality is true and I want to prove it, but just looking at the size is not enough.Eventually to use this reaching an interesting decomposition and using something that results from the initial condition such as $ a + b + c <= 3 $, etc.Intuitively, anyone can look at it and say it's true, RHS looks bigger than LHS, but I need proof. Plus you contradict yourself, because at first you said it seemed true because $ s <p $ and now you use $ s> = p $. I'm sorry if I was mean. – Josie McLaren Mar 17 '21 at 20:53