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Question Number 131465 by bramlexs22 last updated on 05/Feb/21
Let a,b and c be three positive real numbers . Prove that a+b+c ≤ ((a^2 +bc)/(b+c))+((b^2 +ca)/(c+a))+((c^2 +ab)/(a+b))
Leta,bandcbethreepositiverealnumbers.Provethata+b+c⩽a2+bcb+c+b2+cac+a+c2+aba+b
Answered by EDWIN88 last updated on 05/Feb/21
According to the identity ((a^2 +bc)/(b+c))−a=(((a−b)(a−c))/(b+c)) we can change our inequality into the form x(a−b)(a−c)+y(b−a)(b−c)+z(c−a)(c−b)≥0 in which x=(1/(b+c)) , y=(1/(c+a)) ,z=(1/(a+b)) WLOG ,assume a≥b≥c then x≤y≤z
Accordingtotheidentitya2+bcb+c−a=(a−b)(a−c)b+cwecanchangeourinequalityintotheformx(a−b)(a−c)+y(b−a)(b−c)+z(c−a)(c−b)⩾0inwhichx=1b+c,y=1c+a,z=1a+bWLOG,assumea⩾b⩾cthenx⩽y⩽z

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