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Problem 11825

(American Mathematical Monthly, Vol.122, March 2015) Proposed by M. Dinc˘a and S. Radulescu (Romania).

Let E be a normed linear space. Given x1, . . . xn∈ E (with n ≥ 2) such that kxkk = 1 for 1 ≤ k ≤ n and the origin of E is in the convex hull of {x1, . . . , xn}, prove that kx1+ · · · + xnk ≤ n − 2.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Since the origin of E is in the convex hull of {x1, . . . , xn}, it follows that there exist non-negative numbers α1, . . . , αn such that

α1+ · · · + αn = 1 and α1x1+ · · · + αnxn = 0.

We show that αk∈ [0, 1/2] for 1 ≤ k ≤ n. In fact

α1= k − α1x1k = kα2x2+ · · · + αnxnk ≤ α2kx2k + · · · + αnkxnk = α2+ · · · + αn= 1 − α1. which implies that α1≤ 1/2. The other cases are similar. Hence

kx1+ · · · + xnk = kx1+ · · · + xn− 2(α1x1+ · · · + αnxn)k = k(1 − 2α1)x1+ · · · + (1 − 2αn)xnk

≤ (1 − 2α1)kx1k + · · · + (1 − 2αn)kxnk = (1 − 2α1) + · · · + (1 − 2αn) = n − 2.

Finally, we note that right-hand side of the inequality can not be lowered. Take any unit vector x1, let x2= · · · = xn−1= x1and xn = −x1. Then the origin of E is in the convex hull of {x1, . . . , xn} because 12x1+12xn = 0, and kx1+ · · · + xnk = k(n − 2)x1k = n − 2.

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