Menu Close

If-n-1-a-and-b-are-positive-real-numbers-Then-prove-that-a-n-b-n-a-b-a-n-1-b-n-1-2-

Tinku Tara 0 Comments
Pin




Question Number 175815 by Shrinava last updated on 07/Sep/22
If n≥1 a and b are positive real numbers Then prove that: ((a^n + b^n )/(a + b)) ≥ ((a^(n−1) + b^(n−1) )/2)
Ifn⩾1aandbarepositiverealnumbersThenprovethat:a\boldsymboln+b\boldsymbolna+b⩾a\boldsymboln−1+b\boldsymboln−12
Answered by mahdipoor last updated on 07/Sep/22
⇔ 2a^n +2b^n ≥a^n +b^n +ab^(n−1) +ba^(n−1) ⇔ a^n +b^n ≥ab^(n−1) +ba^(n−1) ⇔a^(n−1) (a−b)≥b^(n−1) (a−b) ⇔ { ((⇔^(a−b≥0) a^(n−1) ≥b^(n−1) ⇔ a≥b)),((⇔^(a−b≤0) a^(n−1) ≤b^(n−1) ⇔ a≤b)) :}
⇔2an+2bn⩾an+bn+abn−1+ban−1⇔an+bn⩾abn−1+ban−1⇔an−1(a−b)⩾bn−1(a−b)⇔{⇔a−b⩾0an−1⩾bn−1⇔a⩾b⇔a−b⩽0an−1⩽bn−1⇔a⩽b
Answered by Cesar1994 last updated on 07/Sep/22
(a+b)(a^(n−1) +b^(n−1) )=a^n +b^n +ab^(n−1) +a^(n−1) b ...(1) we proof that a^n +b^n ≥ab^(n−1) +a^(n−1) b...(2) a^n −a^(n−1) b+b^n −ab^(n−1) =a^(n−1) (a−b)+b^(n−1) (b−a) =(a^(n−1) −b^(n−1) )(a−b) without loss of generality, if a≥b>0 ⇒ a^n −a^(n−1) b+b^n −ab^(n−1) =(a^(n−1) −b^(n−1) )(a−b)≥0 ⇒(a+b)(a^(n−1) +b^(n−1) )≤2(a^n +b^n ), using 1 and 2 ⇒((a^(n−1) +b^(n−1) )/2)≤((a^n +b^n )/(a+b)) ■
(a+b)(an−1+bn−1)=an+bn+abn−1+an−1b…(1)weproofthatan+bn⩾abn−1+an−1b…(2)an−an−1b+bn−abn−1=an−1(a−b)+bn−1(b−a)=(an−1−bn−1)(a−b)withoutlossofgenerality,ifa⩾b>0⇒an−an−1b+bn−abn−1=(an−1−bn−1)(a−b)⩾0⇒(a+b)(an−1+bn−1)⩽2(an+bn),using1and2⇒an−1+bn−12⩽an+bna+b◼

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *