Page 24 - MATINF Nr.2

24 L.M. Giugiuc, A. Mirzakhmedov

s
m
m
s
ln 3 ñ (ab) + (bc) + (ca) m s ñ (ab) + (bc) + (ca) s m
ô
ô
Let = s. From Holder’s inequality, ≥ .
2 ln 2 3 3
s
s
(ab) + (bc) + (ca) s
Also, from above, ≥ 1 ⇒
3
s s s m s s s s
ñ ô ñ ô
(ab) + (bc) + (ca) (ab) + (bc) + (ca)
≥ ⇒
3 3
m m m s s s s s
(ab) + (bc) + (ca) (ab) + (bc) + (ca)
ñ ô ñ ô
≥ ⇒
3 3
s
s
m
m
m
s
m
(ab) + (bc) + (ca) ≥ (ab) + (bc) + (ca) ≥ 3 = min {3, 4 } .
The proof is complete. And the problem is closed as well.
Bibliografy
[1] https://cms.math.ca/crux/v43/n3/Solutions 43 3.pdf, page 115.