The SOS method

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
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,

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 )}



  1. well done 🙂

  2. 😀 … It isnt my idea…I only read it..understood it and blogged it….:D 😀

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: