[...]... ƒin™e ( 1 − z 2 + z)2 = 1 + 2z 1 − z2 ≥ 1 ‡e get f (m) ≥ 1 —nd when two of xDyDz —re zero ‡e h—ve f = 1, soW egetfmin = 1 vet z3 z h(z) = + + 2 − 2z 2 , 2 2 e—sy to prove th—t 1 1 h (z) > 0 ⇔ 0 ≤ z < √ andh (z) < 0 ⇔ √ < z ≤ 1 3 3 then ‡e get f (m) ≤ h 1 √ 3 √ 8 3 = , 9 √ √ 8 3 1 8 3 D so ‡e getfmax = when x = y = z = √ W ehavef = 9 9 3 honeF Solution PF ‡hen two of xDyDz —re zero ‡e h—vef = 1D —nd... (b + c) ) a2 + bc a(a − b)(a − c) a2 + bc a (a − b)(a − c)(4 − 2 )≥0 a + bc (a − b)(a − c) ≥ PW essuming ‡vyq a≥b≥c then e—sy to see th—t 4− a ≥0 a2 + bc 4− c ≥0 c2 + ab —nd (c − a)(c − b)(4 − a c ) ≥ 0and( a − b)(a − c)(4 − 2 )≥0 c2 + ab a + bc ‡e h—ve two ™—ses g—se I 4− b2 b ≤0 + ac then (b − c)(b − a)(4 − so this ineq is true g—se P 4− e—sy to see th—t 4− ƒo LHS ≥ (c − b)2 (4 − c2 b2 b )≥0 + ac b