The SOS Theorem

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Joined: Mar 10, 2015

by henderson, Jun 10, 2016, 4:08 PM

$\bf\color{green}Theorem \ (Pham \ King \ Hung) \$ Consider the sum $S=S_a(b-c)^2+S_b(c-a)^2+S_c(a-b)^2.$ Prove that $S\geq 0$ if any of the following take place:
$\color{blue}\Longrightarrow$ $\color{blue}S_a, S_c, S_a+2S_b, S_c+2S_b$ are all non-negative.
$\color{red}\Longrightarrow$ $\color{red}S_b, S_a+S_b$ and $\color{red}S_b+S_c$ are all non-negative when $\color{red}b$ is the median of $\color{red}\{$ $\color{red}a,b,c$ $\color{red}\}.$
$\color{green}\Longrightarrow$ $\color{green}S_b,S_c$ and $\color{green}b^2S_a+a^2S_b$ are all non-negative when $\color{green}a\geq b\geq c.$

$\bf\color{red}Proof.$ For the first part, the $AM-GM$ $(a+c)^2+(2b)^2\geq 4b(a+c)$ yields $(a-c)^2\leq 2(a-b)^2+2(b-c)^2.$ If $S_b\geq 0,$ we are already done; else make the estimate
$\color{blue}S\geq S_c(a-b)^2+S_a(c-b)^2+(2(a-b)^2+2(b-c)^2)S_b=(S_c+2S_b)(a-b)^2+(S_a+2S_b)(c-b)^2\geq 0.$
For the second part, we have
$\color{red}S=S_a(b-c)^2+S_b(c-b+b-a)^2+S_c(a-b)^2=\left(S_a+S_b\right)(b-c)^2+2(c-b)(b-a)S_b+\left(S_c+S_b\right)(a-b)^2\geq0.$
The third part follows from observing that when $a\geq b\geq c$ we have $\frac{a-c}{b-c}\geq \frac{a}{b},$ so that $(a-c)^2\geq \frac{a^2}{b^2}(b-c)^2.$ Hence, we drop the $S_c(a-b)^2$ term to obtain
$\color{green}S\geq S_a(b-c)^2+S_b(a-c)^2\geq S_a(b-c)^2+S_b \left(\frac{a^2(b-c)^2}{b^2}\right)=\left(\frac{b-c}{b}\right)^2(b^2S_a+a^2S_b)\geq 0.$