prove for real x,y and a that (√((x+a)^2 +y^2 ))+(√((x−a)^2 +y^2 ))≥2(√(x^2 +y^2 )) .


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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

	$‘ ‘ ⩾^{″} i s f o r t h e c a s e w h e n t h e 3 p o i n t s$ $a r e c o l i n e a r a n d t h e t r i a n g l e b e c o m e s$ $a s t r a i g h t l i n e . T h i s i s t h e c a s e w h e n$ $a = 0 .$
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