Problem 12163
(American Mathematical Monthly, Vol.127, February 2020) Proposed by T. Speckhofer (Austria).
Let Rnhave the usual dot product and norm. When v= (x1, . . . , xn) ∈ Rn, let Σv = x1+ · · · + xn. Prove
kvk2kwk2≥ (v · w)2+(kvk |Σw| − kwk |Σv|)2 n
for all v, w∈ Rn.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We will show the more general inequality
kvk2kwk2− (v · w)2 kuk2≥ k(w, u)v − (v, u)wk2 where u ∈ Rn.
By letting u = (1, . . . , 1), we have kuk2= n, (v, u) =P v, and (w, u) = P w and we find (kvk2kwk2− (v · w)2) n ≥ k(Σw)v − (Σv)wk2≥ (|Σw| kvk − |Σv| kwk)2 which is equivalent to the given inequality.
If v and w are linearly dependent then kvk2kwk2 = (v · w)2 the inequality holds. We assume now that v and w are linearly independent. Then
u= αv + βw + z where z ⊥ v, z ⊥ w and α, β ∈ R. Moreover
((v, u) = αkvk2+ β(v, w) (w, u) = α(v, w) + βkwk2 . and by solving the linear system we find
α=(v, u)kwk2− (w, u)(v, w)
kvk2kwk2− (v, w)2 and β = (w, u)kvk2− (v, u)(v, w) kvk2kwk2− (v, w)2 . Hence
kvk2kwk2− (v · w)2 kuk2= kvk2kwk2− (v · w)2 (kαv + βwk2+ kzk2)
≥ kvk2kwk2− (v · w)2 (kαv + βwk2)
= kvk2kwk2− (v · w)2 (α2kvk2+ β2kwk2+ 2αβ(v, w))
= (w, u)2kvk2+ (v, u)2kwk2− 2(w, u)(v, u)(v, w)
= k(w, u)v − (v, u)wk2.
