Prove-that-x-y-R-7x-2-6xy-2y-2-x-3-gt-0-

Question Number 11834 by Mr Chheang Chantria last updated on 02/Apr/17

P r o v e t h a t \forall x, y \in R

\Rightarrow 7 x^{2} - 6 x y + 2 y^{2} + x + 3 > 0

Answered by mrW1 last updated on 02/Apr/17

7 x^{2} - 6 x y + 2 y^{2} + x + 3

= {(\frac{3}{\sqrt{2}} x)}^{2} - 2 \times \frac{3}{\sqrt{2}} x \times \sqrt{2} y + {(\sqrt{2} y)}^{2} + {(\frac{1}{2} x)}^{2} + 2 \times \frac{1}{2} x + 1 + [7 - {(\frac{3}{\sqrt{2}})}^{2} - {(\frac{1}{2})}^{2}] x^{2} + (3 - 1)

= {(\frac{3}{\sqrt{2}} x - \sqrt{2} y)}^{2} + {(\frac{1}{2} x + 1)}^{2} + {(\frac{3}{2} x)}^{2} + 2

> 2 > 0

Commented by Mr Chheang Chantria last updated on 03/Apr/17

G o o d s o l u t i o n;)

Answered by ajfour last updated on 02/Apr/17

6 x^{2} - 6 x y + 2 y^{2} + x^{2} + x + 3

= 2 x^{2} [3 - \frac{3 y}{x} + {(\frac{y}{x})}^{2}] + {(x + \frac{1}{2})}^{2} - \frac{1}{4} + 3

= 2 x^{2} [{(\frac{y}{x} - \frac{3}{2})}^{2} - \frac{9}{4} + 3] + {(x + \frac{1}{2})}^{2} + \frac{11}{4}

= 2 x^{2} [\frac{{(2 y - 3 x)}^{2}}{4 x^{2}} + \frac{3}{4}] + {(x + \frac{1}{2})}^{2} + \frac{11}{4}

= \frac{{(2 y - 3 x)}^{2}}{2} + \frac{3 x^{2}}{2} + {(x + \frac{1}{2})}^{2} + \frac{11}{4} > 0

Commented by Mr Chheang Chantria last updated on 03/Apr/17

{that}^{'} s so sweet;)

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 02/Apr/17

7 x^{2} + (1 - 6 y) x + 2 y^{2} + 3 = 0

Δ = {(1 - 6 y)}^{2} - 4 \times 7 (2 y^{2} + 3) =

1 - 12 y + 36 y^{2} - 56 y^{2} - 84 =

- 20 y^{2} - 12 y - 83 = - (20 y^{2} + 12 y + 83)

Δ^{^{'}} = 12^{2} - 4 \times 20 \times 83 < 0

b e c a u s e Δ^{'} < 0, t h e n Δ o n l y h a v e o n e

s i g n t h a t s i m i l a r t o t h e c o e f f i c e n t o f

y^{2} i . e : (- 20) . s o a l w y e s Δ < 0 a n d t h e

7 x^{2} - 6 x y + 2 y^{2} + 3, a n y w e r e h a v e o n e

s i g n t h a t s i m i l a r t o s i g n o f x^{2}, i . e : (+ 7)

s o t h i s p o l y n o m i a l i s p o s i t i v e a n y w e r e .

Commented by mrW1 last updated on 02/Apr/17

g o o d a n d i n t e r e s t i n g p o i n t o f v i e w!

Commented by Mr Chheang Chantria last updated on 03/Apr/17

n i c e s o l u t i o n a n d n i c e e x p l a i n

T h a n k s y o u;)