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### Prove Triangle of Sides with Length...

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Date: 8/1/96 at 21:27:38
From: Anonymous
Subject: Prove Triangle of Sides with Length...

I have a problem for you from the Asian Pacific Mathematics Olympiad
held on March 12, 1996:

Question 5:

Let a, b and c be the lengths of the sides of a triangle. Prove that
the square root of (a+b-c) plus the square root of (b+c-a) plus the
square root of (c+a-b) is equal to or less than the sum of the square
roots of a, b and c. Determine when equality occurs.

Intuitively, equality occurs when the triangle is equilateral. My
teacher has spent weeks on it but has been unable to reach a final
proof.

```

```
Date: 8/2/96 at 11:25:18
From: Doctor Anthony
Subject: Re: Prove Triangle of Sides with Length...

The information that a, b, c are the sides of a triangle has no
significance beyond the fact that all the terms like a+b-c will be
positive (two sides of a triangle are together greater than the third
side).  So we let

a+b-c = p^2
b+c-a = q^2
c+a-b = r^2    From these equations  2a = r^2+p^2
2b = p^2+q^2
2c = q^2+r^2

So a = (r^2+p^2)/2    b = (p^2+q^2)/2     c = (q^2+r^2)/2  and we are

sqrt(a) + sqrt(b) + sqrt(c) >or= p + q + r  ...(1)

Now use the fact that (x^2+y^2)/2 > {(x+y)/2}^2  (See later for proof
of this)

Substituting for a, b, c in expression(1) we have:

sqrt{(r^2+p^2)/2} + sqrt{(p^2+q^2)/2} + sqrt{(q^2+r^2)/2}

>or= sqrt{(r+p)/2}^2 + sqrt{(p+q)/2}^2 + sqrt{(q+r)/2}^2

>or= (1/2){r+p + p+q + q+r}

>or= (1/2){2p + 2q + 2r}

>or= {p + q + r}

And this is the required inequality, i.e.

sqrt(a) + sqrt(b) + sqrt(c) >or= sqrt(a+b-c) + sqrt(b+c-a) +
sqrt(c+a-b)

NOTE:  We must show that (x^2+y^2)/2 > {(x+y)/2}^2  as promised above.

This inequality requires  2(x^2+y^2) > x^2 + 2xy + y^2

x^2 - 2xy + y^2 > 0

(x-y)^2 > 0    and this is always true.
Hence the inequality is true.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Basic Algebra
High School Geometry
High School Triangles and Other Polygons

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