Acute Triangle Inequality

[jay_shark]

Show that for any acute triangle with sides a,b,c , the following inequality holds .

1/(a^2+b^2-c^2) +1/(b^2+c^2-a^2)+1/(a^2+c^2-b^2)>=3/(abc)^2/3

[jay_shark]

Just to clarify things , the rhs should be 3/(abc)^(2/3)

I hate doing math on computer .

[jay_shark]

Here is my solution to the few who may be interested .

First Label 1/(a^2+b^2-c^2)=x
1/(b^2+c^2-a^2)=y
1/(a^2+c^2-b^2)=z

And denote the lhs by W . Then x+y+z=w
Also , note that each of x,y,z are positive quantities from the cosine law since the triangle is acute .

Now (x+y) +(x+z) + (y+z) = 2w
x+y=2b^2/(x^-1*y^-1). Do the same for (x+z),
(y+z)

Notice that we can use the arithmetic/geometric inequality on the denominator of (x+y) .
Since [(x^-1+ y^-1]/2 >= sqrt(x^-1*y^-1)
Which implies that 2b^2/2 >= sqrt(x^-1*y^-1)
or b^4>= x^-1*y^-1 .

But x+y=2b^2/(x^-1*y^-1)>= 2b^2/b^4

Now all together we have that 2w>= 2/b^2 + 2/a^2+ 2/c^2
Divide both sides by two and apply the arithmetic geometric inequality one more time .

W>=1/b^2+1/a^2+1/c^2>=3(1/b^2*1/a^2*1/c^2)^1/3 which is exactly the rhs .

[anisotropy]

Maybe I'm way off here, but can you explain how (a^2+b^2-c^2) does not equal zero? Just by the application of the pythagorean theorem, wouldn't it stand that the denominator of the first fraction is 0?

[Justin A]

[ QUOTE ]
Maybe I'm way off here, but can you explain how (a^2+b^2-c^2) does not equal zero? Just by the application of the pythagorean theorem, wouldn't it stand that the denominator of the first fraction is 0?

[/ QUOTE ]

Pythagorean theorem relates only to right triangles.

[anisotropy]

[ QUOTE ]
Pythagorean theorem relates only to right triangles.

[/ QUOTE ]

And hence, my blatent blunder. Thanks /images/graemlins/blush.gif