prove-for-real-x-y-and-a-that-x-a-2-y-2-x-a-2-y-2-2-x-2-y-2-

Question Number 13929 by ajfour last updated on 25/May/17

p r o v e f o r r e a l x, y a n d a t h a t

\sqrt{{(x + a)}^{2} + y^{2}} + \sqrt{{(x - a)}^{2} + y^{2}} ⩾ 2 \sqrt{x^{2} + y^{2}} .

Answered by mrW1 last updated on 25/May/17

{L e t}^{'} s c o n s i d e r 3 p o i n t s i n c o o r d i n a t e

s y s t e m :

A (x, y)

B (- x, - y)

C (a, 0)

B C = \sqrt{{(x + a)}^{2} + y^{2}}

C A = \sqrt{{(x - a)}^{2} + y^{2}}

B A = \sqrt{{(2 x)}^{2} + {(2 y)}^{2}} = 2 \sqrt{x^{2} + y^{2}}

I n t h e t r i a n g l e Δ A B C w e h a v e

B C + C A ⩾ B A

\Rightarrow \sqrt{{(x + a)}^{2} + y^{2}} + \sqrt{{(x - a)}^{2} + y^{2}} ⩾ 2 \sqrt{x^{2} + y^{2}}

Commented by mrW1 last updated on 25/May/17

Commented by ajfour last updated on 25/May/17

Commented by ajfour last updated on 25/May/17

l e t u s c a l l o r i g i n a s O .

a s y o u s h o w e d A B < A C + B C

2 (A O) < A C + A C^{'}

\Rightarrow A O < \frac{A C + A C^{'}}{2}

s u m o f m e d i a n l e n g t h s < s u m o f s i d e s .

Commented by Joel577 last updated on 25/May/17

when that inequality is equal ?

because the question used “ ⩾^{″} sign

Commented by ajfour last updated on 25/May/17

t h e r e r e m a i n s n o t, a t r i a n g l e!

Commented by mrW1 last updated on 25/May/17

“ ⩾^{″} i s f o r t h e c a s e w h e n t h e 3 p o i n t s

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

a s t r a i g h t l i n e . T h i s i s t h e c a s e w h e n

a = 0 .

Facebook Tweet Pin