# Derive the inequality

1. Nov 23, 2006

### tehno

For a triangle with sides a,b,c and its corresponding circle with radius R:

$$\frac{a^2b^2}{c^2} +\frac{a^2c^2}{b^2}+\frac{b^2c^2}{a^2} \geq 9R^2$$

Last edited: Nov 23, 2006
2. Nov 23, 2006

### Office_Shredder

Staff Emeritus
What is the corresponding circle? Inscribed or circumscribed?

3. Nov 23, 2006

### tehno

Circumscribed of course (usually denoted by capital R).

4. Nov 24, 2006

### tehno

Anybody?
I prooved it in a long and ugly way.
I'd like to see ,if possible,an elegant proof of it.

5. Nov 29, 2006

### murshid_islam

me too.
and tehno, can you please post your proof or at least an outline of it?

6. Dec 2, 2006

### tehno

Nothing particularly smart.
I used substitution:
$$R=\frac{abc}{4P}$$
where P is area of triangle with sides a,b,c.
I went to proove :
$$\frac{ab}{c^2}+\frac{ac}{b^2}+\frac{bc}{a^2}>\frac{9}{4}$$
which I used along the way to proove the original inequality having on mind basic triangle inequality $$a+b>c$$.
After lot of algebraic work I arrived at the original inequality.

But can we somehow make a use something more elegant like a well known :
$$\frac{1}{a^2+b^2+c^2}\geq \frac{1}{9R^2}$$

?

7. Dec 20, 2006

### tehno

It can't be possible you have a proof of it becouse the inequality is invalid!
Some things were odd and by closer inspection I found error in my proof.
The error helped me also to find an obvious counterexample when ineqality doesn't hold.Consider the triangle with following parameters:
$$a=b=1;c=\sqrt{3};R=1$$

8. Jan 23, 2007

### tehno

I will rewrite the expression in a trigonometric form and give a restriction.

$$(sin(\alpha) sin(\beta) cosec(\gamma))^2+(sin(\alpha) cosec(\beta) sin(\gamma))^2+ (cosec(\alpha) sin(\beta) sin(\gamma))^2\geq\frac{9}{4}$$

The restriction is the inequality holds for acute triangles.Now,when I fixed it,
proove the claim.