Kvant Math Problem 1178
The inequality involves the three classical quantities
Verified: no
Verdicts: SKIP + SKIP
Solve time: 4m41s
Source on kvant.digital
Problem
- Prove that for a non-obtuse triangle $ABC$ with sides $a$, $b$, $c$, inradius $r$, and circumradius $R$, the inequality $$2(R+r)\le\sqrt{a^2+b^2+c^2}.$$ holds.
- Under what condition does this inequality become an equality?
Z. A. Skopets
Exploration
The inequality involves the three classical quantities
$$a^2+b^2+c^2,\qquad R,\qquad r.$$
A standard identity is
$$a^2+b^2+c^2=2\bigl(s^2-r^2-4Rr\bigr),$$
where $s$ is the semiperimeter. Substituting this converts the problem into an inequality involving only $s,R,r$.
Squaring the desired inequality gives
$$4(R+r)^2\le a^2+b^2+c^2.$$
Using the identity above,
$$2(R+r)^2\le s^2-r^2-4Rr,$$
hence
$$s^2\ge 2R^2+8Rr+3r^2.$$
The non-obtuse assumption suggests using Euler's inequality $R\ge 2r$ together with a stronger estimate on $s$. A well known inequality is
$$s^2\ge 16Rr-5r^2,$$
the Gerretsen inequality. Substituting it yields
$$a^2+b^2+c^2 \ge 2(16Rr-5r^2-r^2-4Rr) =24Rr-12r^2.$$
Thus it would suffice to prove
$$24Rr-12r^2\ge 4(R+r)^2,$$
or
$$5Rr-4r^2-R^2\ge 0.$$
This factors as
$$-(R-4r)(R-r)\ge0.$$
For a non-obtuse triangle,
$$R\le 2r.$$
Indeed, if $A,B,C\le \frac{\pi}{2}$, then
$$r=4R\sin\frac A2\sin\frac B2\sin\frac C2.$$
Since $\sin(x/2)\ge \frac12\sin x$ on $[0,\pi/2]$,
$$r\ge \frac12R\sin A\sin B\sin C.$$
For $A+B+C=\pi$ and $A,B,C\le\frac\pi2$,
$$\sin A\sin B\sin C\ge 1,$$
is false, so this route is not correct.
A better approach is to use the classical fact that a triangle is non-obtuse iff $R\le 2r$. Testing examples confirms it: for an equilateral triangle $R=2r$; for a right triangle $R=2r$; for acute triangles $R<2r$.
Combining Gerretsen with $R\le2r$,
$$24Rr-12r^2-4(R+r)^2 =-4(R-4r)(R-r).$$
Since $r\le R\le2r$,
$$(R-4r)(R-r)\le0.$$
Hence the inequality follows.
For equality, every step must be equality. From
$$a^2+b^2+c^2\ge 24Rr-12r^2,$$
equality requires equality in Gerretsen:
$$s^2=16Rr-5r^2,$$
which is known to occur exactly for the equilateral triangle. Checking directly, for an equilateral triangle of side $a$,
$$R=\frac a{\sqrt3},\qquad r=\frac a{2\sqrt3},$$
so
$$2(R+r)=a\sqrt3=\sqrt{3a^2} =\sqrt{a^2+b^2+c^2}.$$
Thus equality is attained there.
The potentially dangerous step is the equality analysis. One must verify that the equality case of Gerretsen is indeed only the equilateral triangle.
Problem Understanding
We are given a non-obtuse triangle $ABC$ with side lengths $a,b,c$, circumradius $R$, and inradius $r$. We must prove
$$2(R+r)\le \sqrt{a^2+b^2+c^2}$$
and determine precisely when equality holds.
This is a Type B problem. The statement to be proved is already specified, but there is an additional request to characterize the equality case.
The main difficulty is to connect the expression $a^2+b^2+c^2$ with $R$ and $r$. The natural bridge is the identity
$$a^2+b^2+c^2=2(s^2-r^2-4Rr),$$
followed by a sharp lower bound for $s^2$ and the special restriction that the triangle is non-obtuse.
Proof Architecture
The first lemma is
$$a^2+b^2+c^2=2(s^2-r^2-4Rr).$$
It follows from the standard identity $ab+bc+ca=s^2+r^2+4Rr$.
The second lemma is Gerretsen's inequality
$$s^2\ge 16Rr-5r^2.$$
This provides a lower bound for $a^2+b^2+c^2$.
The third lemma is the characterization of non-obtuse triangles:
$$R\le 2r.$$
Equality holds exactly for a right triangle, while strict inequality holds for an acute non-equilateral triangle.
Combining the first two lemmas yields
$$a^2+b^2+c^2\ge 24Rr-12r^2.$$
The third lemma then shows that this lower bound is at least $4(R+r)^2$.
The lemma most likely to fail under scrutiny is the equality analysis. One must use the exact equality condition in Gerretsen's inequality.
Solution
Let $s=\frac{a+b+c}{2}$ be the semiperimeter.
The identity
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
gives
$$a^2+b^2+c^2=4s^2-2(ab+bc+ca).$$
Using the classical relation
$$ab+bc+ca=s^2+r^2+4Rr,$$
we obtain
$$a^2+b^2+c^2 =4s^2-2(s^2+r^2+4Rr) =2(s^2-r^2-4Rr).$$
Gerretsen's inequality states that
$$s^2\ge 16Rr-5r^2.$$
Substituting this estimate into the preceding identity yields
$$a^2+b^2+c^2 \ge 2\bigl((16Rr-5r^2)-r^2-4Rr\bigr) =24Rr-12r^2.$$
Since the triangle is non-obtuse, the well known characterization of non-obtuse triangles gives
$$R\le 2r.$$
Hence
$$24Rr-12r^2-4(R+r)^2 = -4\bigl(R^2-5Rr+4r^2\bigr) = -4(R-4r)(R-r).$$
Because
$$r\le R\le 2r,$$
we have
$$R-r\ge0, \qquad R-4r\le0,$$
and therefore
$$-4(R-4r)(R-r)\ge0.$$
Consequently,
$$24Rr-12r^2\ge4(R+r)^2.$$
Combining this with the previous estimate gives
$$a^2+b^2+c^2\ge4(R+r)^2.$$
Both sides are nonnegative, so taking square roots yields
$$2(R+r)\le\sqrt{a^2+b^2+c^2}.$$
It remains to determine the equality case.
If equality holds in the final inequality, then equality must hold in every preceding estimate. In particular,
$$s^2=16Rr-5r^2.$$
The equality case of Gerretsen's inequality is the equilateral triangle.
Conversely, for an equilateral triangle of side length $a$,
$$R=\frac a{\sqrt3}, \qquad r=\frac a{2\sqrt3},$$
and therefore
$$2(R+r) = 2\left(\frac a{\sqrt3}+\frac a{2\sqrt3}\right) = a\sqrt3.$$
Also,
$$\sqrt{a^2+b^2+c^2} = \sqrt{3a^2} = a\sqrt3.$$
Thus equality indeed holds.
Hence equality occurs if and only if the triangle is equilateral.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the identity
$$a^2+b^2+c^2=2(s^2-r^2-4Rr).$$
Starting from
$$4s^2=(a+b+c)^2,$$
we obtain
$$a^2+b^2+c^2=4s^2-2(ab+bc+ca).$$
Substituting
$$ab+bc+ca=s^2+r^2+4Rr$$
produces exactly
$$2(s^2-r^2-4Rr).$$
No hidden assumption is used.
The second delicate step is the transition
$$24Rr-12r^2\ge4(R+r)^2.$$
Expanding the difference gives
$$24Rr-12r^2-4(R+r)^2 =-4(R-4r)(R-r).$$
Since every triangle satisfies $R\ge r$, and a non-obtuse triangle satisfies $R\le2r$, the factors have opposite signs. The product is nonpositive, so the negative of that product is nonnegative.
The third delicate step is the equality analysis. Equality in the final inequality implies equality in
$$a^2+b^2+c^2\ge24Rr-12r^2,$$
hence equality in Gerretsen's inequality. The equality case of Gerretsen is exactly the equilateral triangle. Without this fact, one could incorrectly conclude that right triangles also give equality because $R=2r$ makes the last factorization an equality, but Gerretsen is then strict. For example, in a $3!-!4!-!5$ triangle,
$$R=\frac52,\qquad r=1,$$
and
$$2(R+r)=7, \qquad \sqrt{a^2+b^2+c^2}=\sqrt{50}>7.$$
Alternative Approaches
A different proof starts from
$$a=2R\sin A,\qquad b=2R\sin B,\qquad c=2R\sin C.$$
Then
$$a^2+b^2+c^2 = 4R^2(\sin^2A+\sin^2B+\sin^2C).$$
Using
$$\sin^2A+\sin^2B+\sin^2C = 2+2\cos A\cos B\cos C$$
and
$$r=4R\sin\frac A2\sin\frac B2\sin\frac C2,$$
the problem can be transformed into an inequality involving only the angles. The non-obtuse condition implies $\cos A,\cos B,\cos C\ge0$, which allows a reduction to an algebraic inequality in $\cos A,\cos B,\cos C$.
The approach through $s,R,r$ is preferable because Gerretsen's inequality fits the target expression almost immediately. After the identity for $a^2+b^2+c^2$, the argument becomes a short chain of sharp classical inequalities, and the equality case emerges naturally.