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Question Number 12040 by Mr Chheang Chantria last updated on 10/Apr/17
Prove that ∀x,y∈R ⇒ 1+ ((x^2 +y^2 +xy)/3) ≥ x+y
Provethat∀x,y∈R⇒1+x2+y2+xy3⩾x+y
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Apr/17
x^2 +y^2 +xy≥3x+3y−3⇒ x^2 −2x+1+y^2 −2y+1+xy−x−y+1≥0 (x−1)^2 +(y−1)^2 +x(y−1)−(y−1)≥0 (x−1)^2 +(y−1)^2 +(x−1)(y−1)≥0 this iniquality,is valid for evrey x&y∈R.■ (A^2 +B^2 +AB≥0 ∀A,B∈R.)
x2+y2+xy⩾3x+3y−3⇒x2−2x+1+y2−2y+1+xy−x−y+1⩾0(x−1)2+(y−1)2+x(y−1)−(y−1)⩾0(x−1)2+(y−1)2+(x−1)(y−1)⩾0thisiniquality,isvalidforevreyx&y∈R.◼(A2+B2+AB⩾0∀A,B∈R.)
Commented by Mr Chheang Chantria last updated on 11/Apr/17
Nice solution. Thanks you
Nicesolution.Thanksyou
Answered by ajfour last updated on 10/Apr/17
let x+y=u and x−y=v ⇒ xy=(1/4)(u^2 −v^2 ) consider : f(x,y)=1+((x^2 +y^2 +xy)/3)−(x+y) = 1+(((x+y)^2 −xy)/3) − (x+y) = 1+(1/3){ u^2 −(1/4)(u^2 −v^2 )}−u =1+ (1/3){(3/4)u^2 +(v^2 /4) }−u = (u^2 /4)−u+1+(v^2 /(12)) = ((u/2)−1)^2 +(v^2 /(12)) ≥ 0 or 1+((x^2 +y^2 +xy)/3)−(x+y) = (((x+y)/2)−1)^2 +(((x−y)^2 )/(12)) ≥ 0 .
letx+y=uandx−y=v⇒xy=14(u2−v2)consider:f(x,y)=1+x2+y2+xy3−(x+y)=1+(x+y)2−xy3−(x+y)=1+13{u2−14(u2−v2)}−u=1+13{34u2+v24}−u=u24−u+1+v212=(u2−1)2+v212⩾0or1+x2+y2+xy3−(x+y)=(x+y2−1)2+(x−y)212⩾0.
Commented by Mr Chheang Chantria last updated on 11/Apr/17
Good Solution. Thanks you
GoodSolution.Thanksyou

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