Derive the inequality

tehno

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



[tex]\frac{a^2b^2}{c^2} +\frac{a^2c^2}{b^2}+\frac{b^2c^2}{a^2} \geq 9R^2 [/tex]

Office_Shredder

What is the corresponding circle? Inscribed or circumscribed?

tehno

Circumscribed of course (usually denoted by capital R).

tehno

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

murshid_islam

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

tehno

Nothing particularly smart.
I used substitution:
[tex]R=\frac{abc}{4P}[/tex]
where P is area of triangle with sides a,b,c.
I went to proove :
[tex]\frac{ab}{c^2}+\frac{ac}{b^2}+\frac{bc}{a^2}>\frac{9}{4}[/tex]
which I used along the way to proove the original inequality having on mind basic triangle inequality [tex]a+b>c[/tex].
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 :
[tex]\frac{1}{a^2+b^2+c^2}\geq \frac{1}{9R^2}[/tex]

?

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:
[tex]a=b=1;c=\sqrt{3};R=1[/tex]

tehno

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

[tex](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}[/tex]

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