(Van Khea’s inequality): Let , ; ( = 1, 2, … , ; = 1, 2, … , ) be positive real numbers. Suppose that , , … are positive real numbers satisfying − − ⋯ − = 1. Then, we have:
∑ ,∏ ∑ ,≤ ,∏ ,
If − = 1 so we have
,,
+ ,,
+ ⋯ + ,,
≥ , + , + ⋯ + ,, + , + ⋯ + ,
If = ; = = ⋯ = = 1 we have
∑ ,∏ ∑ , ≤ ,∏ ,
: Let , , be positive real numbers satisfy + + = 1. Prove that:
√ + + √ + + √ + ≥ √3√ + 1 ; 2 ≤ ∈ ; ≥ 2
Proof:We have:
a√a + kb + b
√b + kc + c√c + ka = a
√a + kab + b√b + kbc + c
√c + kca≥ (a + b + c)
a + b + c + k(ab + bc + ca) = 1a + b + c + k(ab + bc + ca)
⇒ a√a + kb + b
√b + kc + c√c + ka ≥ 1
(a + b + c) + (k − 2)(ab + bc + ca)⇒ a
√a + kb + b√b + kc + c
√c + ka ≥ 11 + (k − 2)(ab + bc + ca)
But for , , ≥ 0 we have
ab + bc + ca ≤ 13 (a + b + c) ⇒ 1 + (k − 2)(ab + bc + ca)≤ 1 + 13 (k − 2)(a + b + c) = 1 + 13 (k − 2) = 1 + k3⇒ 1 + (k − 2)(ab + bc + ca) ≤ k + 13⇒ 1
1 + (k − 2)(ab + bc + ca) ≥ √3√k + 1
⇒ a√a + kbp + b
√b + kcp + c√c + kap ≥ √3
√k + 1 ; ∀ ≥ 2