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已知a,b,c > 0 且 a^2 + b^2 + c^2 =1


求 ab/c + bc/a + ca/b 的最小值

請問這題該怎麼下手 謝謝

算幾
(ab/c)^2 + (bc/a)^2 ≧ 2b^2
(bc/a)^2 + (ca/b)^2 ≧ 2c^2
(ca/b)^2 + (ab/c)^2 ≧ 2a^2
(ab/c)^2 + (bc/a)^2 + (ca/b)^2 ≧ a^2 + b^2 + c^2 = 1


(ab/c + bc/a + ca/b)^2
= (ab/c)^2 + (bc/a)^2 + (ca/b)^2 + 2(a^2 + b^2 + c^2)
= (ab/c)^2 + (bc/a)^2 + (ca/b)^2 + 2
≧ 3

ab/c + bc/a + ca/b ≧ √3

thepiano 寫:算幾
(ab/c)^2 + (bc/a)^2 ≧ 2b^2
(bc/a)^2 + (ca/b)^2 ≧ 2c^2
(ca/b)^2 + (ab/c)^2 ≧ 2a^2
(ab/c)^2 + (bc/a)^2 + (ca/b)^2 ≧ a^2 + b^2 + c^2 = 1


(ab/c + bc/a + ca/b)^2
= (ab/c)^2 + (bc/a)^2 + (ca/b)^2 + 2(a^2 + b^2 + c^2)
= (ab/c)^2 + (bc/a)^2 + (ca/b)^2 + 2
≧ 3

ab/c + bc/a + ca/b ≧ √3

小弟試了好幾種其他方法都不行,看來只有用鋼琴兄這招才能解決問題!