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]
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] ?
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]
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.