SOS and Smoothing is the habit of writing a symmetric inequality's difference as a sum of squares, or smoothing variables toward equality. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
Many symmetric inequalities are proved by rewriting the difference of the two sides as a sum of squares (so it is visibly $\ge 0$), or by 'smoothing' — nudging unequal variables toward each other to decrease (or increase) the expression until the equality case is reached.
Why It Works
1
Expand the right-hand side: $\tfrac12\big[(a-b)^2+(b-c)^2+(c-a)^2\big] = \tfrac12\big[2a^2+2b^2+2c^2-2ab-2bc-2ca\big]$, which equals $a^2+b^2+c^2-ab-bc-ca$ — so the identity is exact.
2
A sum of real squares is always nonnegative, so $a^2+b^2+c^2 \ge ab+bc+ca$ for all real $a,b,c$, with equality exactly when $a-b=b-c=c-a=0$, i.e. $a=b=c$.
3
The general SOS strategy: to prove a symmetric $f(a,b,c)\ge 0$, try to write $f = S_a(b-c)^2 + S_b(c-a)^2 + S_c(a-b)^2$ with coefficients $S_a,S_b,S_c$ (often depending on $a,b,c$); if these coefficients are nonnegative on the relevant domain the inequality follows.
4
Smoothing is the complementary idea: if replacing two unequal variables by their average (or by some equalizing move) keeps the constraint and weakly decreases the side you want to bound, then the extremum occurs when the variables are equal — which is the SOS equality case. The two viewpoints both certify that equality lives at $a=b=c$.
Worked Examples
Example 1
For real numbers $a,b,c$, prove $a^2+b^2+c^2 \ge ab+bc+ca$, and determine when equality holds.
1
Move everything to one side: the claim is $a^2+b^2+c^2-ab-bc-ca \ge 0$.
2
Apply the SOS identity: $a^2+b^2+c^2-ab-bc-ca = \tfrac12\big[(a-b)^2+(b-c)^2+(c-a)^2\big]$.
3
Each squared term is $\ge 0$, so their half-sum is $\ge 0$; hence $a^2+b^2+c^2 \ge ab+bc+ca$.
4
Equality requires $(a-b)^2=(b-c)^2=(c-a)^2=0$ simultaneously, i.e. $a=b=c$ — the smoothing endpoint where the three variables have been driven together.
Answer:
$a^2+b^2+c^2 \ge ab+bc+ca$ for all real $a,b,c$, with equality iff $a=b=c$.
Contest level (smoothing)
For nonnegative reals with $a + b + c = 3$, show $ab + bc + ca \le 3$, using a smoothing argument.
1
Fix $c$ and the value of $a + b$ (so $a+b+c=3$ still holds), and vary $a, b$. Then $ab + bc + ca = ab + c(a+b)$, in which only the $ab$ term moves.
2
With $a + b$ fixed, $ab = \tfrac14\bigl[(a+b)^2 - (a-b)^2\bigr]$ is largest exactly when $a = b$ (shrinking $|a-b|$ increases it). So smoothing $a, b$ toward each other weakly INCREASES the expression.
3
Repeating the move on each unequal pair drives all three variables together, so the maximum occurs at $a = b = c = 1$.
4
At $a=b=c=1$: $ab+bc+ca = 3$. Hence for all admissible $a,b,c$, $ab+bc+ca \le 3$, with equality iff $a=b=c=1$. (This matches $ab+bc+ca \le \tfrac{(a+b+c)^2}{3} = 3$.)
The expression is symmetric, so assume WLOG $a \ge b \ge c \ge 0$ (this ordering is the SOS/Schur analogue of smoothing).
2
Group the first two terms: $a(a-b)(a-c) + b(b-a)(b-c) = (a-b)\bigl[a(a-c) - b(b-c)\bigr]$. Here $a - b \ge 0$, and $a(a-c) \ge b(b-c)$ because $a \ge b \ge 0$ and $a - c \ge b - c \ge 0$ (larger factors throughout). So this group is $\ge 0$.
3
The last term is $c(c-a)(c-b) = c\,(a-c)(b-c) \ge 0$, since $c \ge 0$, $a - c \ge 0$, and $b - c \ge 0$.
4
Both pieces are nonnegative, so the whole sum is $\ge 0$. Equality requires $a=b=c$, or two equal and the third $0$ (e.g. $a=b,\ c=0$).
Answer:
Schur's inequality holds, with equality iff $a=b=c$ or (two equal, third $=0$).
Going Deeper
The SOS method seeks $f = S_a(b-c)^2 + S_b(c-a)^2 + S_c(a-b)^2$; when the coefficients $S_a,S_b,S_c$ aren't all nonnegative, the SOS criterion (e.g. $S_b, S_b+S_a, S_b+S_c \ge 0$ under an ordering) salvages many otherwise-stubborn symmetric inequalities.
Smoothing / the 'equal-variable' and tangent-line tricks are how olympiad solvers locate and prove the equality case; SOS then certifies it rigorously without calculus — both are mainstays of IMO/USAMO inequality problems.
Pitfall: smoothing only works if the move RESPECTS the constraint and is monotone in the right direction; near a boundary the extremum can sit at $c=0$ (a 'corner'), not at $a=b=c$. Schur's equality at $(t,t,0)$ is exactly such a boundary case — assuming the optimum is interior is a classic error.
Spot the Signal
Use it on symmetric inequalities where equality holds when all variables are equal.
You can describe the hard part as writing a symmetric inequality's difference as a sum of squares, or smoothing variables toward equality, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by rearranging the difference into $\sum(\cdot)^2 \ge 0$, or nudging two variables toward their mean.
Name the relation that makes SOS and Smoothing legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
A smoothing step that changes the constraint, or an SOS form that isn't actually nonnegative.
Applying SOS and Smoothing because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Push the variables toward equality, or expose the squares that were hiding.
Try it on:
Prove a symmetric olympiad inequality via sum-of-squares or smoothing.