Prove that if a + b + c = 1 then ∑ c y c 1 a 2 + b 2 ≤ 9 2 2 My try:since a 2 + b 2 ≥ 2 2 ( a + b )it suffices to prove that ∑ c y c 1 a + b ≤ 9 2 But since Cauchy-Schwarz inequality we have 2 ∑ c y c ( a + b ) ∑ c y c 1 a + b ≥ 9or ∑ c y c 1 a + b ≥ 9 2 so this try can't works.so How to prove ( 1 )

Question

Prove that if   a  +  b  +  c  =  1 then       ∑          c      y      c            1                  a        2            +              b        2              ≤            9              2              2  My try:since            a      2        +          b      2        ≥                    2            2        (  a  +  b  )it suffices to prove that      ∑          c      y      c                  1              a        +        b              ≤            9      2      But since Cauchy-Schwarz inequality we have  2      ∑          c      y      c        (  a  +  b  )      ∑          c      y      c                  1              a        +        b              ≥  9or      ∑          c      y      c                  1              a        +        b              ≥            9      2      so this try can&#039;t works.so How to prove   (  1  )

Sydnee Villegas · Accepted Answer

It&#039;s wrong. Try   a  =  b  →      0    +

rigliztetbf · Accepted Answer

It is false. Try   a  =      9    10    ,    b  =  c  =      1    20

Prove that if a + b + c = 1 then <munder> <mo movablelimits="false">∑<

Answered question

Answer & Explanation

New Questions in Pre-Algebra