IMO 1961 Problem 2
The statement asks for a lower bound on $a^2+b^2+c^2$ in terms of the area $S$ of a triangle.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 11m24s
Problem
Let a,b, and c be the lengths of a triangle whose area is S. Prove that
$a^2 + b^2 + c^2 \ge 4S\sqrt{3}$
In what case does equality hold?
Exploration
The statement asks for a lower bound on $a^2+b^2+c^2$ in terms of the area $S$ of a triangle. Since the area is involved, the first formulas to test are Heron's formula and the trigonometric area formulas
$$S=\frac12 ab\sin C, \qquad S=\sqrt{s(s-a)(s-b)(s-c)},$$
where $s=\frac{a+b+c}{2}$.
The inequality resembles a comparison between a quadratic expression in the side lengths and an area term. Since the right-hand side is homogeneous of degree $2$, scaling preserves the form of the inequality, so an extremal configuration may be highly symmetric. The equilateral triangle is the first candidate for equality. If $a=b=c=t$, then
$$a^2+b^2+c^2=3t^2, \qquad S=\frac{\sqrt3}{4}t^2,$$
hence
$$4S\sqrt3 = 4\cdot \frac{\sqrt3}{4}t^2\cdot \sqrt3 = 3t^2.$$
Equality holds for the equilateral triangle, so the constant is plausible.
A first attempt is to use Heron's formula directly, but this introduces a fourth-degree expression and may obscure equality conditions. A second attempt is to write
$$S=\frac12 ab\sin C \le \frac12 ab,$$
then use
$$a^2+b^2\ge 2ab.$$
This yields
$$a^2+b^2+c^2\ge 2ab\ge 4S,$$
which is weaker than the desired inequality because $\sqrt3>1$.
A more promising direction is to exploit symmetry. Since
$$S=\frac12 ab\sin C,$$
the inequality becomes
$$a^2+b^2+c^2\ge 2\sqrt3,ab\sin C.$$
The law of cosines gives
$$c^2=a^2+b^2-2ab\cos C,$$
so
$$a^2+b^2+c^2 = 2(a^2+b^2)-2ab\cos C.$$
The inequality
$$a^2+b^2\ge 2ab$$
suggests replacing $a^2+b^2$ by $2ab$, which gives
$$a^2+b^2+c^2 \ge 4ab-2ab\cos C = 2ab(2-\cos C).$$
It remains to compare
$$2-\cos C \quad\text{and}\quad \sqrt3\sin C.$$
This reduces to a trigonometric inequality:
$$2-\cos C\ge \sqrt3\sin C.$$
Rewriting,
$$2\ge \cos C+\sqrt3\sin C = 2\cos(C-\pi/3).$$
Since cosine does not exceed $1$, the inequality follows immediately. Equality requires simultaneously $a=b$, equality in the trigonometric bound, and compatibility with the law of cosines. This should force $C=\pi/3$, hence an equilateral triangle.
The delicate point is the equality analysis. Every equality condition from every inequality used must be tracked and intersected carefully.
Problem Understanding
We are given a triangle with side lengths $a,b,c$ and area $S$. The task is to prove a universal inequality relating the sum of the squares of the side lengths to the area:
$$a^2+b^2+c^2\ge 4S\sqrt3,$$
and to determine precisely when equality occurs.
This is a Type C problem because it asks for an extremal relation and requests the equality case. The quantity
$$\frac{a^2+b^2+c^2}{S}$$
is scale invariant, so the problem asks for the smallest possible value of this ratio over all triangles.
The mathematical objects involved are side lengths, triangle area, and trigonometric relations among sides and angles. The main difficulty is that the area depends on both side lengths and angles, so a naive estimate such as
$$S\le \frac12 ab$$
does not recover the sharp constant $\sqrt3$. A successful argument must exploit both algebraic symmetry and a sharp trigonometric estimate that captures the equilateral equality case.
The expected answer is that equality holds precisely for an equilateral triangle. Symmetry suggests this because the inequality is completely symmetric in $a,b,c$, and extremal geometric inequalities for triangles often attain equality at the most symmetric configuration.
Proof Architecture
We will prove the statement through two lemmas.
Lemma 1
For any triangle with angle $C$ opposite side $c$,
$$a^2+b^2+c^2 \ge 2ab(2-\cos C).$$
This follows from the law of cosines together with the inequality
$$a^2+b^2\ge 2ab.$$
Lemma 2
For every angle $C\in(0,\pi)$,
$$2-\cos C\ge \sqrt3\sin C,$$
with equality if and only if
$$C=\frac{\pi}{3}.$$
This follows from rewriting
$$\cos C+\sqrt3\sin C = 2\cos\left(C-\frac{\pi}{3}\right),$$
and using the bound $\cos x\le 1$.
The hardest direction is the equality analysis, because equality in the final inequality requires equality in every intermediate inequality simultaneously. The most likely place for an error is forgetting that equality in
$$a^2+b^2\ge 2ab$$
forces $a=b$.
Solution
We denote by $C$ the angle opposite the side $c$. Since the area of the triangle is
$$S=\frac12 ab\sin C,$$
the desired inequality becomes
$$a^2+b^2+c^2\ge 2\sqrt3,ab\sin C.$$
We first derive a lower bound for $a^2+b^2+c^2$ using the law of cosines.
Lemma 1
For every triangle,
$$a^2+b^2+c^2 \ge 2ab(2-\cos C).$$
Proof
The law of cosines gives
$$c^2=a^2+b^2-2ab\cos C.$$
Hence
$$a^2+b^2+c^2 = 2(a^2+b^2)-2ab\cos C.$$
Since
$$(a-b)^2\ge 0,$$
we obtain
$$a^2+b^2\ge 2ab.$$
Substituting this estimate yields
$$a^2+b^2+c^2 \ge 2(2ab)-2ab\cos C = 2ab(2-\cos C).$$
Equality holds if and only if
$$a=b.$$
∎
This establishes a lower bound expressed only through $ab$ and the angle $C$; replacing $a^2+b^2$ directly by $2ab$ is indispensable because weaker estimates lose the sharp constant.
Lemma 2
For every angle $C\in(0,\pi)$,
$$2-\cos C\ge \sqrt3\sin C,$$
and equality holds if and only if
$$C=\frac{\pi}{3}.$$
Proof
We rewrite the expression
$$\cos C+\sqrt3\sin C = 2\cos\left(C-\frac{\pi}{3}\right).$$
Since
$$\cos x\le 1$$
for all real $x$,
$$\cos C+\sqrt3\sin C \le 2.$$
Rearranging gives
$$2-\cos C\ge \sqrt3\sin C.$$
Equality holds precisely when
$$\cos\left(C-\frac{\pi}{3}\right)=1.$$
Because
$$0<C<\pi,$$
this occurs exactly for
$$C=\frac{\pi}{3}.$$
∎
This establishes the sharp trigonometric estimate; replacing the cosine bound by a rough inequality would not identify the equality case.
We now combine the lemmas. From Lemma 1 and Lemma 2,
$$a^2+b^2+c^2 \ge 2ab(2-\cos C) \ge 2ab\sqrt3\sin C.$$
Using
$$S=\frac12 ab\sin C,$$
we obtain
$$a^2+b^2+c^2 \ge 2ab\sqrt3\sin C = 4S\sqrt3.$$
It remains to determine when equality holds. Equality in the final inequality requires equality in both lemmas simultaneously. Lemma 1 gives
$$a=b,$$
while Lemma 2 gives
$$C=\frac{\pi}{3}.$$
Applying the law of cosines,
$$c^2 = a^2+b^2-2ab\cos\frac{\pi}{3} = a^2+a^2-2a^2\cdot \frac12 = a^2.$$
Hence
$$c=a.$$
Since $a=b$, we conclude
$$a=b=c.$$
Thus the triangle is equilateral.
Conversely, if the triangle is equilateral with side length $t$, then
$$a^2+b^2+c^2=3t^2,$$
and
$$S=\frac{\sqrt3}{4}t^2.$$
Hence
$$4S\sqrt3 = 4\cdot \frac{\sqrt3}{4}t^2\cdot \sqrt3 = 3t^2 = a^2+b^2+c^2.$$
So equality indeed holds.
$$\boxed{a^2+b^2+c^2\ge 4S\sqrt3,\text{ with equality if and only if the triangle is equilateral}.}$$
Verification of Key Steps
The first delicate step is the estimate
$$a^2+b^2\ge 2ab.$$
An independent derivation starts from
$$(a-b)^2=a^2+b^2-2ab.$$
Since a square is nonnegative,
$$a^2+b^2-2ab\ge 0.$$
Equality occurs exactly when
$$a=b.$$
A careless argument may overlook the equality condition and incorrectly claim that any isosceles triangle suffices.
The second delicate step is the trigonometric inequality
$$2-\cos C\ge \sqrt3\sin C.$$
Rearranging,
$$2\ge \cos C+\sqrt3\sin C.$$
Writing the right-hand side in amplitude-phase form gives
$$\cos C+\sqrt3\sin C = 2\cos\left(C-\frac{\pi}{3}\right).$$
Since cosine never exceeds $1$, the maximum is exactly $2$, attained only at
$$C=\frac{\pi}{3}.$$
A careless derivation might bound
$$\sin C\le 1,$$
which would produce
$$2-\cos C\ge 1,$$
far too weak to prove the sharp constant.
The third delicate step is the equality analysis. Equality in a chain
$$X\ge Y\ge Z$$
requires both
$$X=Y \quad\text{and}\quad Y=Z.$$
Hence both
$$a=b \quad\text{and}\quad C=\frac{\pi}{3}$$
must hold. Missing one condition would incorrectly admit non-equilateral triangles.
Alternative Approaches
A substantially different proof uses Heron's formula. Writing
$$S^2=s(s-a)(s-b)(s-c), \qquad s=\frac{a+b+c}{2},$$
one may apply the arithmetic mean-geometric mean inequality:
$$s,\ s-a,\ s-b,\ s-c \le \left(\frac{2s}{4}\right)^4 = \left(\frac{s}{2}\right)^4.$$
Hence
$$S^2 \le \frac{s^4}{16}, \qquad S\le \frac{s^2}{4}.$$
Meanwhile,
$$(a+b+c)^2 \le 3(a^2+b^2+c^2),$$
so
$$s^2 = \frac{(a+b+c)^2}{4} \le \frac34(a^2+b^2+c^2).$$
Combining these inequalities gives
$$S \le \frac{3}{16}(a^2+b^2+c^2),$$
which rearranges to the desired inequality. Equality in both estimates forces
$$a=b=c.$$
The main approach is preferable because it is shorter and isolates the sharp constant through a single trigonometric identity, making the equality case completely transparent.