Olympiad MethodsDifficulty 70-1001 tagged problems

Muirhead's Inequality

Muirhead's Inequality is the habit of comparing symmetric homogeneous sums through majorization in olympiad inequalities. 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.

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The Key Result

$$\text{If } (a_1,\dots,a_n) \succ (b_1,\dots,b_n) \text{ (majorization), then } \sum_{\mathrm{sym}} x_1^{a_1}\cdots x_n^{a_n} \;\ge\; \sum_{\mathrm{sym}} x_1^{b_1}\cdots x_n^{b_n}$$

For nonnegative reals, the symmetric sum with the more "spread out" exponent vector dominates the one with a more "balanced" vector, provided the first exponent vector majorizes the second.

Why It Works

1
Order each exponent tuple in decreasing order. The tuple $(a_i)$ majorizes $(b_i)$ — written $(a)\succ(b)$ — when the two tuples have the same total ($\sum a_i = \sum b_i$) and every leading partial sum of $(a)$ is at least that of $(b)$: $\sum_{i\le k} a_i \ge \sum_{i\le k} b_i$ for all $k$.
2
The symmetric sum $\sum_{\mathrm{sym}}$ runs over all $n!$ permutations of the variables, keeping the exponent multiset fixed, so each side is a symmetric polynomial in the $x_i$.
3
Majorization can be realized by a finite sequence of "Robin Hood" transfers that move a little exponent from a larger coordinate to a smaller one; each single transfer only decreases the symmetric sum, by an AM–GM / SOS step on the two affected terms.
4
Chaining these transfers proves $\sum_{\mathrm{sym}} x^{a} \ge \sum_{\mathrm{sym}} x^{b}$, with equality iff $(a)=(b)$ as multisets or all $x_i$ are equal. Note the inequality requires the $x_i \ge 0$ and equal exponent sums; without majorization it can fail.

Worked Examples

Example 1

For positive reals $a,b,c$, prove $2(a^3+b^3+c^3) \ge a^2b + a^2c + b^2a + b^2c + c^2a + c^2b$.

1
Write both sides as symmetric sums. The left side is $\sum_{\mathrm{sym}} a^3 b^0 c^0$: over all $3! = 6$ permutations this gives $2(a^3+b^3+c^3)$, with exponent tuple $(3,0,0)$.
2
The right side is $\sum_{\mathrm{sym}} a^2 b^1 c^0 = a^2b+a^2c+b^2a+b^2c+c^2a+c^2b$, with exponent tuple $(2,1,0)$.
3
Check majorization: both tuples sum to $3$, and the partial sums satisfy $3 \ge 2$ and $3 \ge 3$, so $(3,0,0) \succ (2,1,0)$.
4
Muirhead's inequality then gives $\sum_{\mathrm{sym}} a^3 \ge \sum_{\mathrm{sym}} a^2 b$, i.e. $2(a^3+b^3+c^3) \ge a^2b+a^2c+b^2a+b^2c+c^2a+c^2b$, with equality iff $a=b=c$.

Answer:

$2(a^3+b^3+c^3) \ge a^2b+a^2c+b^2a+b^2c+c^2a+c^2b$, since $(3,0,0)\succ(2,1,0)$.

Contest level

For positive reals $a,b,c$, prove $a^2 + b^2 + c^2 \ge ab + bc + ca$ via Muirhead.

1
Symmetrize. $\sum_{\mathrm{sym}} a^2 b^0 c^0$ runs over all $6$ permutations of exponents $(2,0,0)$: each of $a^2,b^2,c^2$ appears twice, so it equals $2(a^2+b^2+c^2)$.
2
$\sum_{\mathrm{sym}} a^1 b^1 c^0$ over permutations of $(1,1,0)$ equals $2(ab+bc+ca)$.
3
Majorization: both tuples sum to $2$, and partial sums give $2 \ge 1$ and $2 \ge 2$, so $(2,0,0) \succ (1,1,0)$.
4
Muirhead yields $2(a^2+b^2+c^2) \ge 2(ab+bc+ca)$; divide by $2$. Equality iff $a=b=c$.

Answer:

$a^2+b^2+c^2 \ge ab+bc+ca$, equality iff $a=b=c$.

Olympiad / Challenge

For positive reals $a,b,c$, prove $a^4 + b^4 + c^4 \ge abc\,(a+b+c)$.

1
The right side $abc(a+b+c) = a^2bc + ab^2c + abc^2$ has exponent tuple $(2,1,1)$; symmetrizing, $\sum_{\mathrm{sym}} a^2 b c = 2\,abc(a+b+c)$ (the tuple has a repeat, so its $3$ distinct permutations each appear twice).
2
The left side symmetrizes as $\sum_{\mathrm{sym}} a^4 b^0 c^0 = 2(a^4 + b^4 + c^4)$, exponent tuple $(4,0,0)$.
3
Majorization: both sum to $4$; partial sums give $4 \ge 2$, $4 \ge 3$, $4 \ge 4$, so $(4,0,0) \succ (2,1,1)$.
4
Muirhead gives $2(a^4+b^4+c^4) \ge 2\,abc(a+b+c)$; divide by $2$. Equality iff $a=b=c$.

Answer:

$a^4+b^4+c^4 \ge abc(a+b+c)$, since $(4,0,0)\succ(2,1,1)$.

Going Deeper

Muirhead is the symmetric-sum special case of the Hardy–Littlewood–Pólya majorization theorem $\sum f(a_i) \ge \sum f(b_i)$ for convex $f$ when $(a)\succ(b)$; the Schur-convex viewpoint generalizes it well beyond monomials.
It is a workhorse for olympiad symmetric inequalities (USAMO/IMO shortlist), often after homogenizing a constrained inequality so both sides have equal total degree.
Pitfall: Muirhead needs BOTH a FULLY SYMMETRIC sum AND majorization. It does not apply to cyclic sums (e.g. $\sum a^2 b$ cyclic, not symmetric) or to non-comparable tuples like $(3,1,1,1)$ vs $(2,2,2,0)$ where neither majorizes (partial sums $3,4,5$ vs $2,4,6$ cross) — and it requires $x_i \ge 0$.

Spot the Signal

Look for problems where the key step is comparing symmetric homogeneous sums through majorization in olympiad inequalities.
You can describe the hard part as comparing symmetric homogeneous sums through majorization in olympiad inequalities, 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 identify the olympiad structure that calls for muirhead's inequality, then rewrite the givens around it.
Name the relation that makes Muirhead's Inequality 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

Do not use Muirhead's Inequality just because the surface looks familiar; verify the required condition first.
Applying Muirhead's Inequality because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let muirhead's inequality reveal the structural theorem; then compute only what remains.

Try it on:

Practice a contest problem where the key step is comparing symmetric homogeneous sums through majorization in olympiad inequalities.