# Proving a well-known inequality using S.O.S

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 $a, b, c$.

• is S.O.S. Sum of Squares? Jan 27, 2017 at 10:16
• @S.C.B. Yes, it is. Jan 27, 2017 at 10:17
• Not every nonnegative multivariate polynomial is a sum of squares. Jan 27, 2017 at 10:18
• @quasi in $\sum_{cyc}S_A(b-c)^2$, $S_A$ is not necessarily non-negative. Jan 27, 2017 at 10:20
• 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 $f(u,v,w) = \left(\frac{u}{v}\right)^2+\left(\frac{v}{w}\right)^2+\left(\frac{w}{u}\right)^2-3$ Jan 27, 2017 at 10:33

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: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{1}{6abc}\sum_{cyc}(a-b)^2(3c+a-b)$$

• Very nice. But where are the coefficients $S_A,S_B$? Jan 27, 2017 at 11:01
• @SSepehr I fixed my post. Jan 27, 2017 at 11:09

Here is another SOS (Shortest) $$ab^2+bc^2+ca^2-3abc={\frac {a \left( c-a \right) ^{2} \left({b}^{2}+ ac+cb \right) +b \left( 2\,ab+{c}^{2}-3\,ac \right) ^{2}}{4\,ab+ \left( c-a \right) ^ {2}}} \geqslant 0$$

• It is a nice SOS. Aug 21, 2020 at 2:39

$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3 \Leftrightarrow a^2c + b^2a + c^2b \ge 3abc$$ 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 $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3 = \sum_{cyc}\frac{3b+c-a}{6abc}(c-a)^2.$$
• But the factors $3b+c-a$ are not necessity all non-negative, or am I overlooking something? Jan 27, 2017 at 12:00
• 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. Jan 27, 2017 at 12:07
• Since the inequality is cyclic$,$ you must check when $a\leqslant b \leqslant c$ too! Aug 8, 2020 at 7:00

Because$$:$$ $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} -3=\frac{ab^2 +bc^2 +ca^2 -3abc}{abc}$$ It's enough to prove$$:$$ $$ab^2 +bc^2 +ca^2 -3abc \geqslant 0$$ by SOS.

Let $$\text{P}=2(a+b+c)\cdot (ab^2 +bc^2 +ca^2 -3abc)$$ We have$$:$$ $$\text{P}=a \left( a+2\,b \right) \left( b-c \right) ^{2}+c \left( c+2\,a \right) \left( a-b \right) ^{2}+b \left( b+2\,c \right) \left( c-a \right) ^{2}$$ Let me explain the method$$,$$ let $$\text{P}_{\text{sos}}=\sum \left( {\it QQ}_{{1}}{a}^{2}+{\it QQ}_{{4}}ab+{\it QQ}_{{5}}ac+{\it QQ}_{{2}}{b}^{2}+{\it QQ}_{{6}}bc+{\it QQ}_{{3}}{c}^{2} \right) \left( a-b \right) ^{2}$$ Get an identity give $$\left\{\begin{matrix} -{\it QQ}_{{1}}-{\it QQ}_{{2}}=0&\\2\,{\it QQ}_{{1}}-{\it QQ}_{ {4}}-{\it QQ}_{{6}}=0&\\2\,{\it QQ}_{{2}}-{\it QQ}_{{4}}-{\it QQ}_{{5}}+ 2=0&\\2\,{\it QQ}_{{3}}+{\it QQ}_{{5}}+{\it QQ}_{{6}}-4=0&\\-{\it QQ}_{{1} }-{\it QQ}_{{2}}-2\,{\it QQ}_{{3}}+2\,{\it QQ}_{{4}}+2=0 & \end{matrix}\right.$$

Solve this with $$Q_i \geqslant 0 ,(i=1..6)$$ give us$$:$$ $$\left\{ {\it QQ}_{{1}}=0,{\it QQ}_{{2}}=0,{\it QQ}_{{3}}=1,{\it QQ}_{ {4}}=0,{\it QQ}_{{5}}=2,{\it QQ}_{{6}}=0 \right\}$$

Take it into $$\text{P}_{\text{sos}}$$ give the SOS form. By the same way$$,$$ we have$$:$$ $$\text{P}=\frac{1}{3} \sum (2ab-bc-ca)^2 +\sum 2ab(b-c)^2$$

We have $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3 = \frac{(a-b)^2}{(a+b)b} + \frac{(b-c)^2}{(a+b)c} + \frac{(c-a)^2b}{(a+b)ca}.$$

• +1), Nice SOS expression. Aug 9, 2020 at 2:52
• @tthnew Your SOS is nice as well. Aug 9, 2020 at 2:53
• I found another, see now. Aug 21, 2020 at 2:31
• @tthnew Can you find a simple SOS by your tool: math.stackexchange.com/questions/3799517/… Aug 22, 2020 at 15:13