# $4x^{4} + 4y^{3} + 5x^{2} + y + 1\geq 12xy$ for all positive $x,y$

Prove this inequality: $$4x^4 + 4y^3 + 5x^2 + y + 1 \geq 12xy$$ if $x$ and $y$ are real and positive.

Please, I am a beginner and have no idea how to solve this, so don't use any strange theorems.

• from a graph it looks like it might be enough that $y>0$ – Mirko Dec 8 '14 at 18:31
• Sure, if it is true for $y>0$ then $x<0$ would make RHS negative, while LHS is independent of $sgn(x)$. – Milly Dec 8 '14 at 18:38

I shall use only the inequality $a+b\ge 2\sqrt{ab}.$ We have
$$4x^4+1\ge 2\sqrt{4x^4\cdot 1}=4x^2,$$ $$4y^3+y\ge 2\sqrt{4y^3\cdot y}=4y^2.$$
Then $$4x^4 + 4y^3 + 5x^2 + y + 1\ge 4x^2+4y^2+5x^2\ge 2\sqrt{9x^2\cdot 4y^2}=12xy.$$