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Using $AM-GM$ inequality, it is easy to show for $a,b,c>0$, $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3.$$ However, I can't seem to find an S.O.S form for $a,b,c$ $$f(a,b,c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3 = \sum_{cyc}S_A(b-c)^2 \ge 0.$$

Update:

Please note that I'm looking for an S.O.S form for $a, b, c$, or a proof that there is no S.O.S form for $a, b, c$. Substituting other variables may help to solve the problem using the S.O.S method, but those are S.O.S forms for some other variables, not $a, b, c$.

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    is S.O.S. Sum of Squares? – 2017-01-27
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    @S.C.B. Yes, it is. – 2017-01-27
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    Not every nonnegative multivariate polynomial is a sum of squares. – 2017-01-27
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    @quasi in $\sum_{cyc}S_A(b-c)^2$, $S_A$ is not necessarily non-negative. – 2017-01-27
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    @SSepehr -- If you don't require $S_A$ to be always nonnegative, what use would it be in proving the required inequality? – 2017-01-27
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    @quasi S.O.S is the name for a well-known technique for proving inequalities. Try googling S.O.S technique. – 2017-01-27
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    But if the coefficients multiplying the squares aren't nonnegative, you can't conclude that the sum is greater or equal to zero -- at least not that way. – 2017-01-27
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52576/discussion-between-ssepehr-and-quasi). – 2017-01-27
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    Note that the S.O.S. does not make any assumptions on the sign of $a$, $b$, and $c$. Therefore the S.O.S. would be $\geq 0$ even if $f(a,b,c)<0$. This shows that you cannot prove you statement using this approach, not using the condition $a,b,c>0$. Try $f(u,v,w) = \left(\frac{u}{v}\right)^2+\left(\frac{v}{w}\right)^2+\left(\frac{w}{u}\right)^2-3$ – 2017-01-27

2 Answers 2

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Let $c=\min\{a,b,c\}$. Hence, $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{a}{b}+\frac{b}{a}-2+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-1=\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}\geq0$$

From here we can get a SOS form: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{1}{6abc}\sum_{cyc}(a-b)^2(3c+a-b)$$

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    Very nice. But where are the coefficients $S_A,S_B$? – 2017-01-27
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    @SSepehr I fixed my post. – 2017-01-27
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$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3 \Leftrightarrow a^2c + b^2a + c^2b \ge 3abc$$ So we use these S.O.S forms:

  • $a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a+b+c)\sum_{cyc}(a-b)^2$.
  • $a^3 + b^3 + c^3 - a^2c - b^2a - c^2b = \frac{1}{3}\sum_{cyc}a^3 + a^3 + c^3-3a^2c = \frac{1}{3}\sum_{cyc} (2a+c)(c-a)^2$. Hence, the S.O.S form would be $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3 = \sum_{cyc}\frac{3b+c-a}{6abc}(c-a)^2.$$
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    But the factors $3b+c-a$ are not necessity all non-negative, or am I overlooking something? – 2017-01-27
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    There are some conditions for $S_A,S_B,S_C$ which imply that the inequality holds. one of them is $S_A,S_B,S_C \ge 0$. but if you take $a\ge b\ge c$, saying that $S_A + 2S_B \ge 0$ and $S_C + 2S_B \ge 0$ is also enough. – 2017-01-27