# The SOS method

Fundamental
Squares of all Real Numbers are non-negative
This statement on very deep explorations leads us to this method
given to prove inequalities involving big expression we shall first use algebraic methods to factor it into the following form-
$S_{c} (a - b)^2 + S_{b} (a - c)^2 + S_{a}(b - c)^2 \ge 0$ (we have to prove this)

here note that $S_{a},S_{b},S_{c}$ are only real numbers implying they can be negative.
WLOG : $a\ge b\ge c$
such an assuption is correct for symmetric inequalities but for cyclic inequalities we have to consider a extra case – $c \ge b \ge a$ — (*)
now lets observe some cases.
1.   When One of the coefficients are nonegative
Let $S_{b}$(or Sc or Sa) due to our assumption (*) is
nonegative or due to some given condition .
Because of our assumption we first observe the
$(a - c)^2 \ge (a - b)^2 + (b - c)^2$
multiply the inequality throughout by $S_{b}$ (since it is nonegative and wont switch the inequality signs)
and add $S_{c} (a - b) + S_{a} (b - c)$
thus we finally have to prove that
$S_{c} (a - b)^2 + S_{b} (a - c)^2 + S_{a}(b - c)^2 \ge (S_{c} + S_{b})(a - b)^2 + (S_{a} + S_{b})(b - c)^2$
thus we have vastly reduced our big expression into proving that $S_{b} + S_{c}$ and $S_{b} + S_{c}$ are non-negative [b]if[/b]
$S_{b} \ge 0$

2. When one of the coefficients is not positive
[u][b]A[/b][/u]
Let $S_{b} \le 0$, this again can be due to the assumption (*)or some condition given
We observe another inequality due to our assumption,
$2\cdot (a - b)^2 + 2\cdot (b - c)^2 \ge (a - c)^2$
now again we multiply throughout by $S_{b}$ (note that the inequality sign flips as  $S_{b} \le 0$)
now we have,
$S_{c}(a - b)^2 + S_{b} (a - c)^2 + S_{a}(b - c)^2 \ge (S_{c} + 2S_{b})(a - b)^2 + (2S_{b} + S_{a})(b - c)^2$
thus we are left to prove only that, $S_{c} + 2S_{b} \ge 0$,
$2S_{b} + S_{a} \ge 0$, [b]if[/b] we have $S_{b} \le 0$,

[b][u]B[/u][/b]
Due to our assumption(*) we have the following inequality
$\frac {(a - c)^2}{(b - c)^2} \ge \frac {a}{b}$
Let $Satex S_{b} , S_{c} \ge 0$,
we have,
$S_{b} (a - c)^2 + S_{a} (b - c)^2 = (b - c)^2[S_{b}(\frac {(a - c)^2}{(b - c)^2} + S_{a} ] \ge (b - c)^2(S_{b} \cdot \frac {a^2}{b^2} + S_{a}) = \frac {(b - c)^2}{b^2} \cdot [ (a^2)S_{b} + (b^2)S_{a} ]$
thus we have reduced the inequality to proving $(a^2)S_{b} + (b^2)S_{a} \ge 0$

thus we have discovered the ideas and techniques of the method – Sum Of Squares  😀
thus for let $A$ – expression to be proven
$A \ge 0$ if,

1. $S_{b} , S_{b} + S_{a} , S_{b} + S_{c} \ge 0$

2. $S_{a},S_{c} \ge 0$, $2(S_{b}) + S_{c} , 2(S_{b}) + S_{a} \ge 0$

3.$S_{b} ,S_{c} \ge 0 , a^2(S_{b}) + b^2(S_{a}) \ge 0$

4.$S_{a}, S_{b} , S_{c} \ge 0$

Now For Application ,
although there are various problems that i could possible post now i shall post this famous problem from Iran TST 1996-
Q. prove that for all positive reals $x,y,z \ge 0$,
$\frac {1}{(x + y)^2} + \frac {1}{(y + z)^2} + \frac {1}{(z + x)^2} \ge \frac {9}{4(xy + yz + zx )}$