$4x^4+4y^3+5x^2+y+1 \ge 9x^2+4y^2 \ge 12xy$ QED

Too easy. Easier than problem 1.

i think problem 1 is more easier.

Can you explain with more detail for the first step?

$ 4x^4+1 \ge 4x^2 $ (from AM-GM applied for two numbers) so $ 4x^4+5x^2+1=(4x^4+1)+5x^2 \ge 4x^2+5x^2=9x^2 $

mihaith wrote: $ 4x^4+1 \ge 4x^2 $ (from AM-GM applied for two numbers) so $ 4x^4+5x^2+1=(4x^4+1)+5x^2 \ge 4x^2+5x^2=9x^2 $ And how about the inequality with $y$?We have $4x^4+1+5x^2\ge 4x^2+5x^2=9x^2$ and $4y^3+y\ge 4y^2$.

PiAreSquare asked only for the first step, huricane! Anyway, the inequality with y was very simple.

mihaith wrote: PiAreSquare asked only for the first step, huricane! Anyway, the inequality with y was very simple. Well,the first step is $ 4x^4+4y^3+5x^2+y+1\ge 9x^2+4y^2$.