Note on one inequality and its application in intuitionistic fuzzy sets theory. Part 1

: The inequality µ 1 ν + ν 1 µ ≤ 12 is introduced and proved, where µ and ν are real numbers, for which µ, ν ∈ [0 , 1] and µ + ν ≤ 1 . The same inequality is valid for µ = µ A ( x ) , ν = ν A ( x ) , where µ A and ν A are the membership and the non-membership functions of an arbitrary intuitionistic fuzzy set A over a ﬁxed universe E and x ∈ E . Also, a generalization of the above inequality for arbitrary n ≥ 2 is proposed and proved.


Introduction
The Intuitionistic Fuzzy Sets (IFSs) are introduced by K. Atanassov in [1,2] as follows. Let E be a universal set, µ A , ν A : E → I := [0, 1] be mappings and for each x ∈ E: Then the set is called an IFS. Mappings µ A and ν A are called membership and non-membership functions for the element x ∈ E to the set A ⊆ E.
When for each x ∈ E: the set A is transformed to the ordinary fuzzy (Zadeh's) set [4].

Main results
The main result of the paper is the following Theorem 1.
Theorem 1. Let µ, ν ∈ I be real numbers satisfying inequality Then the inequality Before giving the proof of Theorem 1, we need the following lemma.
Lemma. Let the function f be given by Then function f is strictly concave on interval (0, 1) and also strictly decreasing on the same interval. Also, f (1) = 0 and if we define Proof. Using (5) we obtain: and For x ∈ (0, 1) we have The series ∞ n=0 x n n + 1 converges uniformly on each compact set [δ 1 , δ 2 ], where 0 < δ 1 < δ 2 < 1.
This follows from Weierstrass criterion for uniform convergence of series (see [3]), since converges (from D'Alembert criterion for convergence of series (see [3]).
Then the above mentioned function f is strictly concave on interval (0, 1). Also, and the Lemma is proved.
We must note that Corollary 2 means that for fuzzy sets (4) is always true for µ, ν ∈ I, since and f (1) = 0.
Proof of Theorem 1. From Corollary 1, we have that Theorem 1 is valid for µ + ν = 1.
Let us now look at one generalization of (4) for arbitrary n ≥ 2, proving the following theorem.
Theorem 3. Let n ≥ 2 be an arbitrary integer and x i ∈ (0, 1), i = 1, 2, . . . , n be real numbers, such that Then the inequality holds and the equality is possible if and only if x 1 = x 2 = · · · = 1 n . Proof.
Since the sequence 1 − 1 n n ∞ n=1 is strictly increasing and lim n→+∞ 1 − 1 n n = 1 e , then as a corollary of Theorem 3, we obtain the inequality 1 x is strictly decreasing on (0, 1) (see the Lemma), then Therefore, we have and (33) is proved. Finally, we must mention that if A is a fixed IFS over a universe E, then we can construct the following two new sets These sets are IFSs, because for each x ∈ E: µ A (x) 1 π A (x) ≤ µ A (x) + ν A (x) + π A (x) = 1.

Conclusion
In the second part of the present paper, we will represent a new inequality which one may deduce with the help of (4) and the well-known Young's inequality for product. This new inequality also allows IFS interpretation.