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◆❣✉②❡♥ ❉✉② ❚✉♥❣ ✺✻✼ ◆✐❝❡ ❆♥❞ ❍❛r❞ ■♥❡q✉❛❧✐t✐❡s ❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂ ❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂ ✶ ❚❤✐s ♣r♦❞✉❝t ✐s ❝r❡❛t❡❞ ❢♦r ❡❞✉❝❛t✐♦♥❛❧ ♣✉r♣♦s❡✳ P❧❡❛s❡ ❞♦♥✬t ✉s❡ ✉s❡ ✐t ❢♦r ❛♥② ❝♦♠♠❡❝✐❛❧ ♣✉r♣♦s❡ ✉♥❧❡ss ②♦✉ ❣♦t t❤❡ r✐❣❤t ♦❢ t❤❡ ❛✉t❤♦r✳ P❧❡❛s❡ ❝♦♥t❛❝t ❊♠❛✐❧✿♥❣✉②❡♥❞✉②t✉♥❣✾✹❅❣♠❛✐❧✳❝♦♠ ✷ ✶✳ ❛✮ ✐❢ a, b, c ❛r❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs✱ t❤❡♥ a b + b c + c a ≥ a 2 + 1 b 2 + 1 + b 2 + 1 c 2 + 1 + c 2 + 1 a 2 + 1 . ❜✮▲❡t a, b, c, d ❜❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs✳Pr♦✈❡ t❤❛t a 2 − bd b + 2c + d + b 2 − ca c + 2d + a + c 2 − db d + 2a + b + d 2 − ac a + 2b + c ≥ 0. ❙♦❧✉t✐♦♥✿ ❛✮❇② ❈❛✉❝❤②✲❙❝❤✇❛r③✬s ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡✿ a 2 + b 2 (a 2 + 1) (b 2 + 1) ≥ a 2 + b 2 (ab + 1) = ab a 2 + b 2 + a 2 + b 2 ≥ ab a 2 + b 2 + 2 ⇒ a b + b a = a 2 + b 2 ab ≥ a 2 + b 2 + 2 (a 2 + 1) (b 2 + 1) = a 2 + 1 b 2 + 1 + b 2 + 1 a 2 + 1 ❇② ❈❤❡❜②s❤❡✈✬s ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡ a 2 b 2 = a 2 b 2 + 1 + a 2 b 2 (b 2 + 1) ≥ a 2 b 2 + 1 + b 2 b 2 (b 2 + 1) = a 2 + 1 b 2 + 1 . ❚❤❡r❡❢♦r❡ 1 + a b 2 = 1 + 2 a b + b a + a 2 b 2 ≥ 1 + 2 a 2 + 1 b 2 + 1 + b 2 + 1 a 2 + 1 + a 2 + 1 b 2 + 1 = 1 + a 2 + 1 b 2 + 1 2 . ❚❤❡r❡❢♦r❡ a b + b c + c a ≥ a 2 + 1 b 2 + 1 + b 2 + 1 c 2 + 1 + c 2 + 1 a 2 + 1 ❛s r❡q✉✐r❡✳ ❜✮◆♦t✐❝❡ t❤❛t 2(a 2 − bd) b + 2c + d + b + d = 2a 2 + b 2 + d 2 + 2c(b + d) b + 2c + d = (a − b) 2 + (a − d) 2 + 2(a + c)(b + d) b + 2c + d (1) ❆♥❞ s✐♠✐❧❛r❧②✱ 2(c 2 − db) d + 2a + b + b + d = (c − d) 2 + (c − b) 2 + 2(a + c)(b + d) d + 2a + b (2) ❯s✐♥❣ ❈❛✉❝❤②✲❙❝❤✇❛r③✬s ✐♥❡q✉❛❧✐t②✱✇❡ ❣❡t (a − d) 2 b + 2c + d + (c − d) 2 d + 2a + b ≥ [(a − b) 2 + (c − d) 2 ] (b + 2c + d) + (d + 2a + b) (3) ✸ (a − d) 2 b + 2c + d + (c − b) 2 d + 2a + b ≥ [(a − d) 2 + (c − b) 2 ] 2 (b + 2c + d) + (d + 2a + b) (4) 2(a + c)(b + d) b + 2c + d + 2(a + c)(b + d) d + 2a + b ≥ 8(a + c)(b + d) (b + 2c + d) + (d + 2a + b) (5) ❋r♦♠ ✭✶✮✱✭✷✮✱✭✸✮✱✭✹✮ ❛♥❞ ✭✺✮✱ ✇❡ ❣❡t 2( a 2 − bd b + 2c + d + c 2 − db d + 2a + b ) + b + d ≥ (a + c − b − d) 2 + 4(a + c)(b + d) a + b + c + d = a + b + c + d. ♦r a 2 − bd b + 2c + d + c 2 − db d + 2a + b ≥ a + c − b − d 2 ■♥ t❤❡ s❛♠❡ ♠❛♥♥❡r✱✇❡ ❝❛♥ ❛❧s♦ s❤♦✇ t❤❛t b 2 − ca c + 2d + a + d 2 − ac a + 2b + c ≥ b + d − a − c 2 ❛♥❞ ❜② ❛❞❞✐♥❣ t❤❡s❡ t✇♦ ✐♥❡q✉❛❧✐t✐❡s✱✇❡ ❣❡t t❤❡ ❞❡s✐r❡❞ r❡s✉❧t✳ ❊♥q✉❛❧✐t② ❤♦❧❞s ✐❢ ❛♥❞ ♦♥❧② ✐❢ a = c ❛♥❞ b = d✳ ✷✱ ▲❡t a, b, c ❜❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t a + b + c = 1 Pr♦✈❡ t❤❛t t❤❡ ❢♦❧❧♦✇✐♥❣ ✐♥❡q✉❛❧✐t② ❤♦❧❞s ab 1 − c 2 + bc 1 − a 2 + ca 1 − b 2 ≤ 3 8 ❙♦❧✉t✐♦♥✿ ❋r♦♠ t❤❡ ❣✐✈❡♥ ❝♦♥❞✐t✐♦♥ ❚❤❡ ✐♥❡q✉❛❧✐t② ✐s ❡q✉✐✈❛❧❡♥t t♦ 4ab a 2 + b 2 + 2(ab + bc + ca) ≤ 3 2 ❜✉t ❢r♦♠ ❈❛✉❤② ❙❤✇❛r③ ✐♥❡q✉❛❧✐t② 4ab a 2 + b 2 + 2(ab + bc + ca) ≤ ab a 2 + ab + bc + ca + ab b 2 + ab + bc + ca = ab (a + b)(a + c) + ab (b + c)(a + b) = a(b + c) 2 (a + b)(b + c)(c + a) ❚❤✉s ❲❡ ♥❡❡❞ ♣r♦✈❡ t❤❛t 3(a + b)(b + c)(c + a) ≥ 2 a(b + c) 2 ✇❤✐❝❤ r❡❞✉❝❡s t♦ t❤❡ ♦❜✈✐♦✉s ✐♥❡q✉❛❧✐t② ab(a + b) ≥ 6abc ❚❤❡ ❙♦❧✉t✐♦♥ ✐s ❝♦♠♣❧❡t❡❞✳✇✐t❤ ❡q✉❛❧✐t② ✐❢ ❛♥❞ ♦♥❧② ✐❢ a = b = c = 1 3 ✹ ❖r ❲❡ ❝❛♥ ✉s❡ t❤❡ ❢❛❝t t❤❛t 4ab a 2 + b 2 + 2(ab + bc + ca) ≤ 4ab (2ab + 2ac) + (2ab + 2bc) ≤ ab 2a(b + c) + ab 2b(a + c) = 1 2 b b + c + a a + c = 1 2 b b + c + c b + c = 3 2 ✸✱ ▲❡t a, b, c ❜❡ t❤❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs✳ Pr♦✈❡ t❤❛t 1 + ab 2 + bc 2 + ca 2 (ab + bc + ca)(a + b + c) ≥ 4. 3 (a 2 + ab + bc)(b 2 + bc + ca)(c 2 + ca + ab) (a + b + c) 2 ❙♦❧✉t✐♦♥✿ ▼✉❧t✐♣❧②✐♥❣ ❜♦t❤ s✐❞❡s ♦❢ t❤❡ ❛❜♦✈❡ ✐♥❡q✉❛❧✐t② ✇✐t❤ (a + b + c) 2 ✐t✬s ❡q✉✐✈❛❧❡♥t t♦ ♣r♦✈❡ t❤❛t (a + b + c) 2 + (a + b + c)(ab 2 + bc 2 + ca 2 ) ab + bc + ca ≥ 4. 3 (a 2 + ab + bc)(b 2 + bc + ca)(c 2 + ca + ab) ❲❡ ❤❛✈❡ (a + b + c) 2 + (a + b + c)(ab 2 + bc 2 + ca 2 ) ab + bc + ca = (a 2 + ab + bc)(c + a)(c + b) ab + bc + ca ❇② ✉s✐♥❣ ❆▼✲●▼ ✐♥❡q✉❛❧✐t② ❲❡ ❣❡t (a 2 + ab + bc)(c + a)(c + b) ab + bc + ca ≥ 3. 3 (a 2 + ab + bc)(b 2 + bc + ca)(c 2 + ca + ab)[(a + b)(b + c)(c + a)] 2 ab + bc + ca ❙✐♥❝❡ ✐t✬s s✉❢❢✐❝❡s t♦ s❤♦✇ t❤❛t √ 3. 3 (a + b)(b + c)(c + a) ≥ 2. √ ab + bc + ca ✇❤✐❝❤ ✐s ❝❧❡❛r❧② tr✉❡ ❜② ❆▼✲●▼ ✐♥❡q✉❛❧✐t② ❛❣❛✐♥✳ ❚❤❡ ❙♦❧✉t✐♦♥ ✐s ❝♦♠♣❧❡t❡❞✳ ❊q✉❛❧✐t② ❤♦❧❞s ❢♦r a = b = c ✹✱ ▲❡t a 0 , a 1 , . . . , a n ❜❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t a k+1 −a k ≥ 1 ❢♦r ❛❧❧ k = 0, 1, . . . , n−1. Pr♦✈❡ t❤❛t 1 + 1 a 0 1 + 1 a 1 − a 0 ··· 1 + 1 a n − a 0 ≤ 1 + 1 a 0 1 + 1 a 1 ··· 1 + 1 a n ❙♦❧✉t✐♦♥✿ ❲❡ ✇✐❧❧ ♣r♦✈❡ ✐t ❜② ✐♥❞✉❝t✐♦♥✳ ❋♦r n = 1 ❲❡ ♥❡❡❞ t♦ ❝❤❡❝❦ t❤❛t 1 + 1 a 0 1 + 1 a 1 − a 0 ≤ 1 + 1 a 0 1 + 1 a 1 ✇❤✐❝❤ ✐s ❡q✉✐✈❛❧❡♥t t♦ a 0 (a 1 − a 0 − 1) ≥ 0, ✇❤✐❝❤ ✐s tr✉❡ ❜② ❣✐✈❡♥ ❝♦♥❞✐t✐♦♥✳ ▲❡t 1 + 1 a 0 1 + 1 a 1 − a 0 ··· 1 + 1 a k − a 0 ≤ 1 + 1 a 0 1 + 1 a 1 ··· 1 + 1 a k ✺ ✐t r❡♠❛✐♥s t♦ ♣r♦✈❡ t❤❛t✿ 1 + 1 a 0 1 + 1 a 1 − a 0 ··· 1 + 1 a k+1 − a 0 ≤ ≤ 1 + 1 a 0 1 + 1 a 1 ··· 1 + 1 a k+1 ❇② ♦✉r ❤②♣♦t❤❡s✐s 1 + 1 a 0 1 + 1 a 1 ··· 1 + 1 a k+1 ≥ ≥ 1 + 1 a k+1 1 + 1 a 0 1 + 1 a 1 − a 0 ··· 1 + 1 a k − a 0 ✐❞ ❡st✱ ✐t r❡♠❛✐♥s t♦ ♣r♦✈❡ t❤❛t✿ 1 + 1 a k+1 1 + 1 a 0 1 + 1 a 1 − a 0 ··· 1 + 1 a k − a 0 ≥ ≥ 1 + 1 a 0 1 + 1 a 1 − a 0 ··· 1 + 1 a k+1 − a 0 ❇✉t 1 + 1 a k+1 1 + 1 a 0 1 + 1 a 1 − a 0 ··· 1 + 1 a k − a 0 ≥ ≥ 1 + 1 a 0 1 + 1 a 1 − a 0 ··· 1 + 1 a k+1 − a 0 ⇔ ⇔ 1 a k+1 + 1 a k+1 a 0 1 + 1 a 1 − a 0 ··· 1 + 1 a k − a 0 ≥ ≥ 1 (a k+1 − a 0 )a 0 1 + 1 a 1 − a 0 ··· 1 + 1 a k − a 0 ⇔ ⇔ 1 ≥ 1 a k+1 − a 0 1 + 1 a 1 − a 0 ··· 1 + 1 a k − a 0 ❇✉t ❜② ♦✉r ❝♦♥❞✐t✐♦♥s ❲❡ ♦❜t❛✐♥✿ 1 a k+1 − a 0 1 + 1 a 1 − a 0 ··· 1 + 1 a k − a 0 ≤ ≤ 1 k 1 + 1 1 ··· 1 + 1 k − 1 = 1. ❚❤✉s✱ t❤❡ ✐♥❡q✉❛❧✐t② ✐s ♣r♦✈❡♥✳ ✺✱ ●✐✈❡♥ a, b, c > 0✳ Pr♦✈❡ t❤❛t 3 a 2 + bc b 2 + c 2 ≥ 9. 3 √ abc (a + b + c) ❙♦❧✉t✐♦♥ ✿ ❚❤✐s ✐♥❡q ✐s ❡q✉✐✈❛❧❡♥t t♦✿ a 2 + bc 3 abc(a 2 + bc) 2 (b 2 + c 2 ) ≥ 9 (a + b + c) 3 ❇② ❆▼✲●▼ ✐♥❡q ✱ ❲❡ ❤❛✈❡ a 2 + bc 3 abc(a 2 + bc) 2 (b 2 + c 2 ) = ✻ = a 2 + bc 3 (a 2 + bc)c(a 2 + bc)b(b 2 + c 2 )a ≥ 3(a 2 + bc) sym a 2 b ❙✐♠✐❧❛r❧②✱ t❤✐s ✐♥❡q ✐s tr✉❡ ✐❢ ❲❡ ♣r♦✈❡ t❤❛t✿ 3(a 2 + b 2 + c 2 + ab + bc + ca) sym a 2 b ≥ 9 (a + b + c) 3 a 3 + b 3 + c 3 + 3abc ≥ sym a 2 b ❲❤✐❝❤ ✐s tr✉❡ ❜② ❙❝❤✉r ✐♥❡q✳ ❊q✉❛❧✐t② ❤♦❧❞s ✇❤❡♥ a = b = c ✻✱ ▲❡t a, b, c ❜❡ ♥♦♥♥❡❣❛t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t ab + bc + ca > 0✳ Pr♦✈❡ t❤❛t 1 2a 2 + bc + 1 2b 2 + ca + 1 2c 2 + ab ≥ 2 ab + bc + ca . ❚❤❡ ✐♥❡q✉❛❧✐t② ✐s ❡q✉✐✈❛❧❡♥t t♦ ab + bc + ca 2a 2 + bc ≥ 2, (1) ♦r a(b + c) 2a 2 + bc + bc bc + 2a 2 ≥ 2.(2) ❯s✐♥❣ t❤❡ ❈❛✉❝❤②✲❙❝❤✇❛r③ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡ bc bc + 2a 2 ≥ ( bc) 2 bc(bc + 2a 2 ) = 1.(3) ❚❤❡r❡❢♦r❡✱ ✐t s✉❢❢✐❝❡s t♦ ♣r♦✈❡ t❤❛t a(b + c) 2a 2 + bc ≥ 1.(4) ❙✐♥❝❡ a(b + c) 2a 2 + bc ≥ a(b + c) 2(a 2 + bc) ✐t ✐s ❡♥♦✉❣❤ t♦ ❝❤❡❝❦ t❤❛t a(b + c) a 2 + bc ≥ 2, (5) ✇❤✐❝❤ ✐s ❛ ❦♥♦✇♥ r❡s✉❧t✳ ❘❡♠❛r❦✿ 2ca + bc 2a 2 + bc + 2bc + ca 2b 2 + ca ≥ 4c a + b + c . ✼✱ ▲❡t a, b, c ❜❡ ♥♦♥ ♥❡❣❛t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t ab + bc + ca > 0✳ Pr♦✈❡ t❤❛t 1 2a 2 + bc + 1 2b 2 + ca + 1 2c 2 + ab + 1 ab + bc + ca ≥ 12 (a + b + c) 2 . ❙♦❧✉t✐♦♥✿ ✶✮ ❲❡ ❝❛♥ ♣r♦✈❡ t❤✐s ✐♥❡q✉❛❧✐t② ✉s✐♥❣ t❤❡ ❢♦❧❧♦✇✐♥❣ ❛✉①✐❧✐❛r② r❡s✉❧t ✐❢ 0 ≤ a ≤ min{a, b}✱ t❤❡♥ 1 2a 2 + bc + 1 2b 2 + ca ≥ 4 (a + b)(a + b + c) . ✼ ✐♥ ❢❛❝t✱ t❤✐s ✐s ✉s❡❞ t♦ r❡♣❧❛❝❡❞ ❢♦r ✧♥♦ t✇♦ ♦❢ ✇❤✐❝❤ ❛r❡ ③❡r♦✧✱ s♦ t❤❛t t❤❡ ❢r❛❝t✐♦♥s 1 2a 2 + bc , 1 2b 2 + ca , 1 2c 2 + ab , 1 ab + bc + ca ❤❛✈❡ ♠❡❛♥✐♥❣s✳ ❇❡s✐❞❡s✱ t❤❡ ✐❛❦❡r ❛❧s♦ ✇♦r❦s ❢♦r ✐t✿ 1 2a 2 + bc + 1 2b 2 + ca + 1 2c 2 + ab ≥ 2(ab + bc + ca) a 2 b 2 + abc(a + b + c) ❇✉t ♦✉r ❙♦❧✉t✐♦♥ ❢♦r ❜♦t❤ ♦❢ t❤❡♠ ✐s ❡①♣❛♥❞ ▲❡t a, b, c ❜❡ ♥♦♥ ♥❡❣❛t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t ab + bc + ca > 0✳ Pr♦✈❡ t❤❛t 1 2a 2 + bc + 1 2b 2 + ca + 1 2c 2 + ab + 1 ab + bc + ca ≥ 12 (a + b + c) 2 . ✷✮ ❈♦♥s✐❞❡r ❜② ❆▼✲●▼ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡ 2 a 2 + ab + b 2 (a + b + c) = (2b + a) 2a 2 + bc + (2a + b) 2b 2 + ca ≥ 2 (2a + b)(2b + a) (2a 2 + bc) (2b 2 + ca). ❆♥❞ ❜② ❆▼✲●▼ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡ c 2 (2a + b) 2a 2 + bc + c 2 (2b + a) 2b 2 + ca ≥ 2 c 4 (2a + b)(2b + a) (2a 2 + bc) (2b 2 + ca) ≥ 2c 2 (2a + b)(2b + a) (a 2 + ab + b 2 ) (a + b + c) = 4c 2 a + b + c + 6abc a + b + c c a 2 + ab + b 2 2c 2 a + bc 2 + 2ab 2 + b 2 c 2a 2 + bc = c 2 (2a + b) 2a 2 + bc + c 2 (2b + a) 2b 2 + ca ≥ 4c 2 a + b + c + 6abc a + b + c c a 2 + ab + b 2 = 4 a 2 + b 2 + c 2 a + b + c + 6abc a + b + c c a 2 + ab + b 2 ≥ 4 a 2 + b 2 + c 2 a + b + c + 6abc a + b + c (a + b + c) 2 c (a 2 + ab + b 2 ) = 4 a 2 + b 2 + c 2 ab + c + 6abc ab + bc + ca ⇒ 2a 2 b + 2ab 2 + 2b 2 c + 2bc 2 + 2c 2 a + 2ca 2 2a 2 + bc ✽ = (b + c) + 2c 2 a + bc 2 + 2ab 2 + b 2 c 2a 2 + bc ≥ (b + c) + 4 a 2 + b 2 + c 2 a + b + c + 6abc ab + bc + ca = 8 a 2 + b 2 + c 2 + ab + bc + ca a + b + c − 2 a 2 b + ab 2 ab + bc + ca ⇒ 1 2a 2 + bc + 1 ab + bc + ca ≥ 4 a 2 + b 2 + c 2 + ab + bc + ca (a + b + c) ( (a 2 b + ab 2 )) ≥ 12 (a + b + c) 2 . <=> (a + b)(a + c) 2a 2 + bc + a 2 + bc 2a 2 + bc − 2 ≥ 12(ab + bc + ca) (a + b + c) 2 ❋r♦♠ 2a 2 + 2bc 2a 2 + bc − 3 = bc 2a 2 + bc ≥ 1 ❲❡ ❣❡t a 2 + bc 2a 2 + bc − 2 ≥ 0 ◆♦✇✱ ❲❡ ✇✐❧❧ ♣r♦✈❡ t❤❡ str♦♥❣❡r (a + b)(a + c) 2a 2 + bc ≥ 12(ab + bc + ca) (a + b + c) 2 ❋r♦♠ ❝❛✉❝❤②✲s❝❤❛r③t✱ ❲❡ ❤❛✈❡ (a + b)(a + c) 2a 2 + bc = (a+b)(b+c)(c+a)( 1 (2a 2 + bc)(b + c) ≥ 3(a + b)(b + c)(c + a) ab(a + b) + bc(b + c) + ca(c + a) ❋✐♥❛❧❧②✱ ❲❡ ♦♥❧② ♥❡❡❞ t♦ ♣r♦✈❡ t❤❛t (a + b)(b + c)(c + a) ab(a + b) + bc(b + c) + ca(c + a) ≥ 4(ab + bc + ca) (a + b + c) 2 (a + b + c) 2 ab + bc + ca ≥ 4[ab(a + b) + bc(b + c) + ca(c + a) (a + b)(b + c)(c + a) = 4 − 8abc (a + b)(b + c)(c + a) a 2 + b 2 + c 2 ab + bc + ca + 8abc (a + b)(b + c)(c + a) ≥ 2 ✇❤✐❝❤ ✐s ♦❧❞ ♣r♦❜❧❡♠✳ ❖✉r ❙♦❧✉t✐♦♥ ❛r❡ ❝♦♠♣❧❡t❡❞ ❡q✉❛❧✐t② ♦❝❝✉r ✐❢ ❛♥❞ ✐❢ ♦♥❧② a = b = c, a = b, c = 0 ♦r ❛♥② ❝②❝❧✐❝ ♣❡r♠✉t✐♦♥✳ ✽✱ ▲❡t a, b, c ❜❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t 16(a + b + c) ≥ 1 a + 1 b + 1 c ✳ Pr♦✈❡ t❤❛t 1 a + b + 2(a + c) 3 ≤ 8 9 . ❙♦❧✉t✐♦♥✿ ❚❤✐s ♣r♦❜❧❡♠ ✐s r❛t❤❡r ❡❛s②✳ ❯s✐♥❣ t❤❡ ❆▼✲●▼ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡✿ a + b + 2(c + a) = a + b + c + a 2 + c + a 2 ≥ 3 3 (a + b)(c + a) 2 . ✾ ❙♦ t❤❛t✿ 1 a + b + 2(c + a) 3 ≤ 2 27(a + b)(c + a) . ❚❤✉s✱ ✐t✬s ❡♥♦✉❣❤ t♦ ❝❤❡❝❦ t❤❛t✿ 1 3(a + b)(c + a) ≤ 4 ⇐⇒ 6(a + b)(b + c)(c + a) ≥ a + b + c, ✇❤✐❝❤ ✐s tr✉❡ s✐♥❝❡ 9(a + b)(b + c)(c + a) ≥ 8(a + b + c)(ab + bc + ca) ❛♥❞ 16abc(a + b + c) ≥ ab + bc + ca ⇒ 16(ab + bc + ca) 2 3 ≥ ab + bc + ca ⇐⇒ ab + bc + ca ≥ 3 16 . ❚❤❡ ❙♦❧✉t✐♦♥ ✐s ❝♦♠♣❧❡t❡❞✳ ❊q✉❛❧✐t② ❤♦❧❞s ✐❢ ❛♥❞ ♦♥❧② ✐❢ a = b = c = 1 4 ✳ ✾✱ ▲❡t x, y, z ❜❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t xyz = 1✳ Pr♦✈❡ t❤❛t x 3 + 1 x 4 + y + z + y 3 + 1 y 4 + z + x + z 3 + 1 z 4 + x + y ≥ 2 √ xy + yz + zx. ❙♦❧✉t✐♦♥✿ ❯s✐♥❣ t❤❡ ❆▼✲●▼ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡ 2 (x 4 + y + z)(xy + yz + zx) = 2 [x 4 + xyz(y + z)](xy + yz + zx) = 2 (x 3 + y 2 z + yz 2 )(x 2 y + x 2 z + xyz) ≤ (x 3 + y 2 z + yz 2 ) + (x 2 y + x 2 z + xyz) = (x + y + z)(x 2 + yz) = (x + y + z)(x 3 + 1) x . ✐t ❢♦❧❧♦✇s t❤❛t x 3 + 1 x 4 + y + z ≥ 2x √ xy + yz + zx x + y + z . ❆❞❞✐♥❣ t❤✐s ❛♥❞ ✐t ❛♥❛❧♦❣♦✉s ✐♥❡q✉❛❧✐t✐❡s✱ t❤❡ r❡s✉❧t ❢♦❧❧♦✇s✳ ✶✵✱ ▲❡t a, b, c ❜❡ ♥♦♥♥❡❣❛t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s❛t✐s❢②✐♥❣ a + b + c = √ 5✳ Pr♦✈❡ t❤❛t (a 2 − b 2 )(b 2 − c 2 )(c 2 − a 2 ) ≤ √ 5 ❙♦❧✉t✐♦♥✿ ❋♦r t❤✐s ♦♥❡✱ ❲❡ ❝❛♥ ❛ss✉♠❡ ❲▲❖● t❤❛t c ≥ b ≥ a s♦ t❤❛t ❲❡ ❤❛✈❡ P = (a 2 − b 2 )(b 2 − c 2 )(c 2 − a 2 ) = (c 2 − b 2 )(c 2 − a 2 )(b 2 − a 2 ) ≤ b 2 c 2 (c 2 − b 2 ). ❆❧s♦ ♥♦t❡ t❤❛t √ 5 = a + b + c ≥ b + c s✐♥❝❡ a ≥ 0✳ ◆♦✇✱ ✉s✐♥❣ t❤❡ ❆▼✲●▼ ✐♥❡q✉❛❧✐t② ❲❡ ❤❛✈❡ (c + b) · √ 5 2 − 1 · c 2 · √ 5 2 + 1 b 2 · (c − b) ≤ (c + b) √ 5(b + c) 5 5 ≤ √ 5; ❙♦ t❤❛t ❲❡ ❣❡t P ≤ √ 5✳ ❆♥❞ ❤❡♥❝❡ ❲❡ ❛r❡ ❞♦♥❡✳ ❊q✉❛❧✐t② ❤♦❧❞s ✐❢ ❛♥❞ ♦♥❧② ✐❢ (a, b, c) = √ 5 2 + 1; √ 5 2 − 1; 0 ❛♥❞ ❛❧❧ ✐ts ❝②❝❧✐❝ ♣❡r♠✉t❛t✐♦♥s✳ ✷ ✶✵ [...]... ine ( 1 − z 2 + z)2 = 1 + 2z 1 − z2 ≥ 1 e get f (m) ≥ 1 nd when two of xDyDz re zero e hve f = 1, soW egetfmin = 1 vet z3 z h(z) = + + 2 − 2z 2 , 2 2 esy to prove tht 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 hvef = 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 esy to see tht 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 hve two ses gse I 4− b2 b ≤0 + ac then (b − c)(b − a)(4 − so this ineq is true gse P 4− esy to see tht 4− o LHS ≥ (c − b)2 (4 − c2 b2 b )≥0 + ac b