kf k2kgk2 α2kf k2− β2kgk2
2 ≥ 0

Problem 11667

(American Mathematical Monthly, Vol.119, October 2012) Proposed by Cezar Lupu (Romania).

Let f , g, and h be elements of an inner product space over R, with hf, gi = 0.

(a) Show that hf, f ihg, gihh, hi²≥ 4hg, hi²hh, f i².

(b) Show that hf, f ihh, hihh, f i²+ hg, gihh, hihg, hi²≥ 4hg, hi²hh, f i².

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

If f = 0 or g = 0 then (a) and (b) are trivial. Assume that f 6= 0 and g 6= 0. Let r = h − αf − βg where

α = hf, hi

kf k², β = hg, hi kgk², then hr, f i = hr, gi = 0 and hh, hi = α²kf k²+ β²kgk²+ krk². (a) We have that

hf, f ihg, gihh, hi²− 4hg, hi²hh, f i²= kf k²kgk²(α²kf k²+ β²kgk²+ krk²)²− 4α²kf k⁴β²kgk⁴

≥ kf k²kgk² (α²kf k²+ β²kgk²)²− 4α²β²kf k²kgk²

= kf k²kgk² α²kf k²− β²kgk^2
2

≥ 0.

(b) We have that

hf, f ihh, hihh, f i²+ hg, gihh, hihg, hi²− 4hg, hi²hh, f i²

= (α²kf k²+ β²kgk²+ krk²)(α²kf k⁶+ β²kgk⁶) − 4α²kf k⁴β²kgk⁴

≥ (α²kf k²+ β²kgk²)(α²kf k⁶+ β²kgk⁶) − 4α²β²kf k⁴kgk⁴

≥ (α²kf k⁴+ β²kgk⁴)²− 4α²β²kf k⁴kgk⁴

≥ (α²kf k⁴− β²kgk⁴)²≥ 0

where at the third step we used the fact that, by Cauchy-Schwarz inequality, (α²kf k²+ β²kgk²)(α²kf k⁶+ β²kgk⁶) ≥ (α²kf k⁴+ β²kgk⁴)².