Question

Proof Triangle inequality ?

Answer 1

I NOT UNDERSTAND WHAT IT MEANT CAN ANYONE UNDERSTAND ME

Answer 2

Heya User,

--> Take a triangle with sides --> 'a' , 'b' , 'c'

--> Eh! =_= One can say that :-> 'a' can be equal to 'b' and greater than 'c'
--> a = b > c  { we can write this ^_^ }

Or say --> 'a' can be equal to 'b' equal to 'c'
--> a = b = c <-- we can write this ...

Now, an interestingly stupid case =_= :->
--> We take a triangle where a is greater than b, but lesser than 'c'

:-> a > b  and a < c 
---> b < a < c   { We can write it like this }

Eh! Also, we can write any triangle length in a nearly similar way =_=
---> I write a general form :-> a ≥ b ≥ c > 0

Note :-> They can't be '0' ri8 ... or else, we wud be dealing with a straight line =_= 

So, we consider this --> a ≥ b ≥ c > 0

=> a ≥ b  || a ≥ c 
=> a + c > b  || a + b > c   --> { again =_= Side length cannot be 0 }

--> However, a + b + c > 0 

Put all three in the Heron's formula -->
--> Since the area of a Δ is always +ve ... if you're living on earth that is, 
we have :->
--> ( a + b + c ) ( a + b - c ) ( a + c - b ) ( b + c - a ) > 0 

^ ^ -- we write the above by heron's formula ...

But, ( a + b + c ) > 0 || and ( a + b - c ) > 0 || ( a + c - b ) > 0 
=> k ( b + c - a ) > 0 --> where 'k' is a +ve real no....

But, positive integer multiplied by -ve would be a contradiction to the above inequality 

Hence, we have --> ( b + c - a ) > 0 and we're done 

TRIANGLE INEQUALITY --> 

---> a + b > c || a + c > b || b + c > a ^_^