Problem 11972
(American Mathematical Monthly, Vol.124, April 2017) Proposed by Yun Zhang (China).
Let r be the radius of the sphere inscribed in a tetrahedron whose exscribed spheres have radii r1, r2, r3, and r4. Prove
r(√3r1+√3r2+√3r3+√3r4) ≤ 2√3r1r2r3r4.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let us consider an n-simplex in Rn for n ≥ 2. We show that
r
n+1
X
k=1
√nrk ≤ (n − 1)n v u u t
n+1
Y
k=1
rk,
where r is the radius of the inscribed sphere and r1, r2, . . . , rn+1are the radii of the n + 1 exscribed spheres. The required inequality can be obtained by setting n = 3.
We have that
rPn+1
k=1
√nrk
n
q Qn+1
k=1rk
=
n+1
X
k=1
n
s Y
j6=k
r rj ≤
n+1
X
k=1
1 n
X
j6=k
r rj = r
n+1
X
k=1
1
rk = n − 1
where in the second step we applied the AM-GM inequality and in the last step we used the known relation
r
n+1
X
k=1
1
rk = n − 1.
Remark For the sake of completeness, we give here a proof of the above known relation (see for example A.A. Toda, Radii of the inscribed and escribed spheres of a simplex, Int. J. Geom. 3, No.
2, 5-13 (2014)).
Let S = {P1, P2, . . . , Pn+1} be the vertices of the n-simplex, and let Sk = S \ {Pk} be the vertices of the face opposite to the vertex Pk. Then, for 1 ≤ k ≤ n + 1,
conv({Ik} ∪ Sk) ∪
n+1
[
j=1
conv({I} ∪ Sj) = conv({Ik} ∪ S) = [
j6=k
conv({Ik} ∪ Sj)
which implies, by computing the n-dimensional volumes (the sets are pairwise disjoint), rk|Sk|
n +
n+1
X
j=1
r|Sj|
n =X
j6=k
rk|Sj|
n =⇒ r
rk = 1 − 2|Sk| Pn+1
j=1|Sj|. Hence
r
n+1
X
k=1
1 rk
=
n+1
X
k=1
1 − 2|Sk| Pn+1
j=1|Sj|
!
= (n + 1) − 2 = n − 1.
