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USAJMO 2026 (Problem 1)

Let

a,b,c

be distinct positive integers such that

ab+c=c^2

. Prove that

(a-b)^2\ge 4c

.

Core

Algebraic Identities

Major

Properties of Inequalities
Optimization by Equalization

Prove that

(a-b)^2\ge4c\iff a+b\ge2c.

Proof:We have

(a-b)^2=(a+b)^2-4ab=(a+b)^2-4(c^2-c)=(a+b)^2-4c(c-1).

(a-b)^2\ge4c\iff(a+b)^2-4c(c-1)\ge4c

\iff(a+b)^2\ge4c^2

\iff a+b\ge2c.

Assume, for a contradiction, that

a+b<2c

.

a+b<2c\implies a+b\le2c-1.

The maximum value of

ab

occurs when

a

and

b

are as close as possible.Since

a,b,c

are distinct positive integers, neither

a

nor

b

can equal

c

. Hence

ab\le(c-2)(c+1)=c^2-c-2.

From the given condition,

ab=c^2-c\le c^2-c-2.

0\le-2\qquad\text{✗}

This is a contradiction.

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