Can acute triangles satisfy this inequality involving sides and radius?

[tehno]

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



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

 

[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]

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

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

 

[tehno]

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

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 :
$$\frac{1}{a^2+b^2+c^2}\geq \frac{1}{9R^2}$$

?

 

[tehno]

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

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:
$$a=b=1;c=\sqrt{3};R=1$$

 

[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.