Problem 11075
(American Mathematical Monthly, Vol.111, April 2004) Proposed by G. Trenkler (Germany).
Let a, b, and c be complex numbers. Show that
pa2+ b2+ c2
≤ max {|a| + |b|, |b| + |c|, |a| + |c|} .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We consider the complex anti-symmetric matrix
M =
0 a b
−a 0 c
−b −c 0
whose eigenvalues are 0, ±√
a2+ b2+ c2. Then the inequality follows from the Gerschgorin Theorem.
However we give here a direct proof.
Let λ = √
a2+ b2+ c2 (one of the squareroots) and let x = (x1, x2, x3) 6= 0 be a corresponding eigenvector. Since M x = λx then
ax2+ bx3= λx1
−ax1+ cx3= λx2
−bx1− cx2= λx3
and
|λ||x1| ≤ |a||x2| + |b||x3| ≤ (|a| + |b|) · maxi=2,3|xi| ≤ R · maxi=1,2,3|xi|
|λ||x2| ≤ |a||x1| + |c||x3| ≤ (|a| + |c|) · max
i=1,3|xi| ≤ R · max
i=1,2,3|xi|
|λ||x3| ≤ |b||x1| + |c||x2| ≤ (|b| + |c|) · maxi=1,2|xi| ≤ R · maxi=1,2,3|xi|
(1)
where R = max {|a| + |b|, |b| + |c|, |a| + |c|}. Therefore
|λ| · max
i=1,2,3|xi| ≤ R · max
i=1,2,3|xi|.
Since max
i=1,2,3|xi| > 0 then
pa2+ b2+ c2
=λ≤ R = {|a| + |b|, |b| + |c|, |a| + |c|} .
Note that the equality holds if and only if at least two of the numbers a, b and c are zero. If the is only one zero, say c, then
λ=
pa2+ b2
=p|a2+ b2| ≤p|a|2+ |b|2<|a| + |b| = R.
Now assume that a, b and c are all different from zero and that λ = R. Then all the inequalities in (??) become equalities and it follows that
|x1| = |x2| = |x3| 6= 0 and |a| + |b| = |b| + |c| = |a| + |c| = R, that is |a| = |b| = |c| 6= 0 which yields the contradiction
√3|a| = λ = R = 2|a|.
