I found an inequality and I have been stuck for days trying to solve it. I tried everything but nothing I try seems to work. The inequality is:
given $a,b,c{\gt}0$ and that $a^2+b^2+c^2=3$
prove that ${ab\over c}+{ac\over b}+{bc\over a}{\ge}3$
Can somebody please tell me how this to solve this problem?
Assuming $a,b,c$ are positive, it is enough to show the homogeneous: $$\left(\sum_{cyc} \frac{ab}{c}\right)^2 \geqslant 3(a^2+b^2+c^2)$$
$$\iff (a^2b^2+b^2c^2+c^2a^2)^2\geqslant 3(abc)^2(a^2+b^2+c^2)$$
With $x=a^2, y=b^2, z=c^2$, this is the same as $$x^2y^2+y^2z^2+z^2x^2 \geqslant x^2yz+y^2zx+x^2zy$$ which is true by Muirhead or Rearrangement.