## imqwerty one year ago another ques..

1. imqwerty

l,m,n are positive real numbers such that$l^3+m^3=n^3$prove that$l^2+m^2-n^2>6(n-l)(n-m)$

2. MrNood

I am not sure about this - but I was under the impression that there were NO solutions for that equation (isn't this Fermat's last theorem problem?)

3. ganeshie8

Fermat's last theorem was about "integer" solutions

4. mathmath333

$$\color{white}{bookmark}$$

5. AbdullahM

Didn't @kainui solve this yesterday?

6. imqwerty

thas ws not the proper solution :)

7. MrNood

@ganeshie8 ah yes - sorry my mistake

8. imqwerty

ok :) heres the solution-

10. imqwerty

( ͡° ͜ʖ ͡°)=found

11. ParthKohli

That's a crazy solution. You can't really ever normally think of this.

12. imqwerty

solution obtained by normal thinking - u can write that inequality like -$l^2+m^2-n^2-6(n-l)(n-m)$$7n^2-6(l+m)n-(l^2+m^2-6lm)<0$ x=7c^2 y=-6(l+m)n z=-(l^2 +m^2 -6lm) so u have to prove x+y+z<0 u can also observe that x ,y, z are not all equal x>0 , y<0 use the identity- $x^3+y^3+z^3-3xyz=\frac{ 1 }{ 2 }(x+y+z)[(x-y)^2 +(y-z)^2+(z-x)^2]$ but its enough to prove that x^2 + y^3+z^3 -3xyz<0 substitute x, y,z :/ i didn't tell this cause i hate the tedious calculations it involves after some simplification this will reduce to-$-l^2m^2(129l^2+129m^2-254lm)<0$ notice that this thing does is not having n cause i substituted n^3=m^3+l^3 while simplifying and clearly $129l^2 +129m^2 -254lm=129(l-m)^2+4lm>0$ hence the results follow ...

13. ParthKohli

haha normal thinking

14. imqwerty

:D