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I posted this yesterday $4x^{4} + 4y^{3} + 5x^{2} + y + 1\geq 12xy$ for all positive $x,y$ and I think I got a solution , Here it goes ,

this is the problem , if $x$ and $y$ are real positive numbers , I need to prove

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

but $x^2 + y^2 \ge 2xy $, so $6x^2 + 6y^2 \ge 12xy $, from this if

$4x^4 + 4y^3 + 5x^2 + y + 1 \ge 6x^2 + 6y^2$ then the given inequality holds

let's try this $4x^4 + 4y^3 \ge 6x^2 + 6y^2$

$4x^4 + 4y^3 / (6x^2 + 6y^2) \ge 1 $

$ 4x^4 + 4y^3 /(6x^2 + 6y^2) - 1 \ge 0 $

$4x^4 + 4y^3 - 6x^2 - 6y^2 /(6x^2 + 6y^2) \ge 0$ , divide both sides by the numerator

$1/(6x^2 + 6y^2) \ge 0 $

since this is true the inequality $4x^4 + 4y^3 \ge 6x^2 + 6y^2$ holds , so from this we can say that

$4x^4 + 4y^3 + 5x^2 + y + 1 \ge 4x^4 + 4y^3 $

$5x^2 + y +1 \ge 0 $

, this is obviously true and thus the above inequality holds but something tells me thing is wrong , I believe I proved this to be the case for all natural numbers but I intended to prove this equality for all positive real numbers .

Do tell me where i've messed up

Thanks

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The inequality $4x^4 + 4y^3 \ge 6x^2 + 6y^2$ is wrong already for $x=y=1$. So when you divide both sides by the numerator, you don’t obtain neither $1/(6x^2 + 6y^2) \ge 0 $ nor anything tree.

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