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For $a,b,c>0.$ Prove$:$ $${\frac {ab}{ \left( a+b \right) ^{2}}}+{\frac {bc}{ \left( b+c \right) ^{2}}}+{\frac {ac}{ \left( c+a \right) ^{2}}}+\,{\frac { \left( a+b \right) \left( b+c \right) \left( c+a \right) }{16abc}}\geqslant \frac{5}{4}$$ AM-GM kills it easy, but I think it's hard to get SOS$,$ I can't!

If $c=\min\{a,b,c\},$ we obtain the following by Maple$:$ enter image description here However it's ugly. So I wish another SOS.

PS: This inequality is from Nguyen Viet Hung.

There is the AM-GM proof here: https://www.facebook.com/groups/1486244404996949/permalink/2695082927446418/

So I don't need the AM-GM proof.

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Yes, SOS helps.

We need to prove that $$\frac{\prod\limits_{cyc}(a+b)}{16abc}-\frac{1}{2}\geq\sum_{cyc}\left(\frac{1}{4}-\frac{ab}{(a+b)^2}\right)$$ or $$\frac{\sum\limits_{cyc}c(a-b)^2}{16abc}\geq\sum_{cyc}\frac{(a-b)^2}{4(a+b)^2}$$ or $$\sum_{cyc}(a-b)^2\left(\frac{1}{ab}-\frac{4}{(a+b)^2}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)^4}{ab(a+b)^2}\geq0.$$

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    $\begingroup$ WOW, very nice and simple SOS! $\endgroup$
    – NKellira
    Aug 19, 2020 at 7:20
  • $\begingroup$ I still can't understand why me software does not give SOS for this. $\endgroup$
    – NKellira
    Aug 19, 2020 at 7:24
  • $\begingroup$ Yeah, and I have upvote and accept it for you. $\endgroup$
    – NKellira
    Aug 19, 2020 at 7:28
  • $\begingroup$ @tthnew There are many ways to write a polynomial in the SOS form. Just sometimes human make it a bit of better than computer. $\endgroup$ Aug 19, 2020 at 7:30
  • $\begingroup$ Yes, this time my computer does not work. NguyenHuyen_AG give the longer SOS than you by computer. $\endgroup$
    – NKellira
    Aug 19, 2020 at 7:31

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