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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

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

G o o d S o l u t i o n . T h a n k s y o u