If any point $O$ in the interior of a triangle ABC. Prove that AB $+$ BC $+$ CA $>$ OA $+$ OB $+$ OC.

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Solution

$\begin{matrix} I n Δ P B O B P + P O > O B A d d i n g O C a n b o t h s i d e s w e g e t, B P + P O + O C > O B + O C B P + P C > O B + O C B P + P C > O B + O C \to (i) [P O + O C = P C] I n Δ A P C A P + A C > P C A d d i n g B P o n b o t h s i d e s, w e g e t A P + A C + B P > P C + B P \to (i i) [A P + P B = A B] f r o m (i) & (i i) A B + A C > B P + P C > O B + O C A B + A C > O B + O C \to (i i i) s i m i l a r l y B C + A C > O B + O A \to (i v) A B + B C > O A + O C \to (v) a d d i n g e q u a t i o n (3) (4) (5) w e g e t 2 (A B + B C + C A) > 2 (O A + O B + O C) A B + B C + C A > O A + O B + O C \end{matrix}$

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Similar questions

. O is any point in the interior of a triangle ABC. Prove that

1. AB + AC > OB + OC

2.AB + BC + CA > OA + OB + OC

3.OA + OB + OC > 1/2(AB + AC + BC)

O

A B C

O A + O B + O C > 1 / 2 (A B + B C + A C)

O is any point in the interior of $Δ A B C .$

Prove that

$(i) A B + A C > O B + O C$

$(i i) A B + B C + C A > O A + O B + O C$

$(i i i) O A + O B + O C > \frac{1}{2} (A B + B C + C A)$

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>

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O A + O B + O C > \frac{1}{2} (A B + B C + C A)

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