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Monday, January 19, 2015

2015/008) show that if x,y,z are 3 sides of a triangle so are \dfrac{1}{x+y},\dfrac{1}{x+z},\dfrac{1}{y+z}

we need to show that for arbitrary sides of length x,y z

\dfrac{1}{x+z} + \dfrac{1}{y+z}\gt \dfrac{1}{x+y}

or (y+z)(x+y) + (x+z)(x+y) \ge (x+z)(x+z)
or simlifying
x^2+y^2 + xy + yz + xz \ge z^2


now
x^2+y^2 + xy + yz + xz = z(x+y) + x^2+y^2 + xy
\gt z(x+y)
\gt z^2 as x+y \gt z

hence proved

I have taken the problem from http://mathhelpboards.com/challenge-questions-puzzles-28/lengths-three-segments-triangle-14063.html#post66760 where I provided the above solution


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