Prove that ∀x,y∈R ⇒ 1+ ((x^2 +y^2 +xy)/3) ≥ x+y


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Question Number 12040 by Mr Chheang Chantria last updated on 10/Apr/17

$Prove that \forall x, y \in R$ $\Rightarrow 1 + \frac{x^{2} + y^{2} + x y}{3} ⩾ x + y$
	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Apr/17

	$x^{2} + y^{2} + x y ⩾ 3 x + 3 y - 3 \Rightarrow$ $x^{2} - 2 x + 1 + y^{2} - 2 y + 1 + x y - x - y + 1 ⩾ 0$ ${(x - 1)}^{2} + {(y - 1)}^{2} + x (y - 1) - (y - 1) ⩾ 0$ ${(x - 1)}^{2} + {(y - 1)}^{2} + (x - 1) (y - 1) ⩾ 0$ $t h i s i n i q u a l i t y, i s v a l i d f o r e v r e y x & y \in R . ◼$ $(A^{2} + B^{2} + A B ⩾ 0 \forall A, B \in R .)$
	Commented by Mr Chheang Chantria last updated on 11/Apr/17

	$N i c e s o l u t i o n . T h a n k s y o u$
	Answered by ajfour last updated on 10/Apr/17
	$let x+y=u and x−y=v ⇒ xy=(1/4)(u^2 −v^2 ) consider : f(x,y)=1+((x^2 +y^2 +xy)/3)−(x+y) = 1+(((x+y)^2 −xy)/3) − (x+y) = 1+(1/3){ u^2 −(1/4)(u^2 −v^2 )}−u =1+ (1/3){(3/4)u^2 +(v^2 /4) }−u = (u^2 /4)−u+1+(v^2 /(12)) = ((u/2)−1)^2 +(v^2 /(12)) ≥ 0 or 1+((x^2 +y^2 +xy)/3)−(x+y) = (((x+y)/2)−1)^2 +(((x−y)^2 )/(12)) ≥ 0 .$
	$l e t x + y = u a n d x - y = v$ $\Rightarrow x y = \frac{1}{4} (u^{2} - v^{2})$ $c o n s i d e r :$ $f (x, y) = 1 + \frac{x^{2} + y^{2} + x y}{3} - (x + y)$ $= 1 + \frac{{(x + y)}^{2} - x y}{3} - (x + y)$ $= 1 + \frac{1}{3} {u^{2} - \frac{1}{4} (u^{2} - v^{2})} - u$ $= 1 + \frac{1}{3} {\frac{3}{4} u^{2} + \frac{v^{2}}{4}} - u$ $= \frac{u^{2}}{4} - u + 1 + \frac{v^{2}}{12}$ $= {(\frac{u}{2} - 1)}^{2} + \frac{v^{2}}{12} ⩾ 0$ $o r$ $1 + \frac{x^{2} + y^{2} + x y}{3} - (x + y)$ $= {(\frac{x + y}{2} - 1)}^{2} + \frac{{(x - y)}^{2}}{12} ⩾ 0 .$
	Commented by Mr Chheang Chantria last updated on 11/Apr/17

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