Proving $\sum {\frac {ab}{ \left( a+b \right) ^{2}}}+{\frac {\prod \left( a+b \right) }{16abc}}\geq \frac{5}{4}$

inequalitysum-of-squares-methodsummationsymmetric-polynomials

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.

Best Answer

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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