(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