let $x,y$ be positive real numbers. prove that $ 4x^4+4y^3+5x^2+y+1\geq 12xy $
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Tags: inequalities, inequalities unsolved
01.01.2015 08:02
$4x^4+4y^3+5x^2+y+1 \ge 9x^2+4y^2 \ge 12xy$ QED
01.01.2015 08:40
Too easy. Easier than problem 1.
01.01.2015 09:04
i think problem 1 is more easier.
01.01.2015 19:42
Can you explain with more detail for the first step?
01.01.2015 20:28
$ 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 $
01.01.2015 20:31
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 $4y^3+y\ge 4y^2$.
01.01.2015 22:30
PiAreSquare asked only for the first step, huricane! Anyway, the inequality with y was very simple.
01.01.2015 22:41
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$.