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

Question

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 $a, b, c$.

Answer 1

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

Answer 2

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

Answer 3

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$:$ $$\text{P}=\frac{1}{3} \sum (2ab-bc-ca)^2 +\sum 2ab(b-c)^2$$

Answer 4

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

Answer 5

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