Complete square to show that $(x-2y)^4+64xy+16 \geq 0$

Question

I want to prove that

$f(x,y) = (x-2y)^4+64xy+16 \geq 0$

Using only algebraic methods (I guess it can be solved via completing the square).

I didn't manage to do so. Any help will be appreciated.

Thank you in advance.

Answer 1

Write $t = x-2y$ so $x=t+2y$ and now we have $$\begin{align} f(t,y) &= t^4+64y(t+2y)+16 \\ &= 128y^2+64yt +(t^4+16) \end{align}$$ This is quadratic function on $y$. It discriminant is $$\begin{align} D&= 64^2t^2 -4\cdot 128 (t^4 +16)\\ & = -128\cdot 4 (t^4-8t^2+16)\\ &= -128\cdot 4(t^2-4)^2\\ &\leq 0\end{align} $$

and we are done.

Answer 2

You can use also AM-GM inequality $a+b\geq 2\sqrt{ab}$: $$(x-2y)^4+ 16 \geq 8\cdot (x-2y)^2$$

then $$\begin{align} f(x,y) &\geq 8\cdot (x-2y)^2 +64 xy \\ &= 8\cdot (x^2-4xy + 4y^2 +8xy) \\& =8(x^2+4xy +4y^2) \\ & =8(x+2y)^2 \\ &\geq 0 \end{align} $$

Answer 3

You can complete the square.

$f(x,y) = (x-2y)^4+64xy+16$

Note that $4xy = x^2 + 4y^2 - (x-2y)^2 $

So $f(x,y) = (x-2y)^4 + 8 [x^2 + 4y^2 - (x-2y)^2] + 32xy + 16$

$ = (x-2y)^4 - 8 (x-2y)^2 + 16 + 8(x^2+4y^2 + 4xy)$

$ = [(x-2y)^2 - 4]^2 + 8 (x+2y)^2 \geq 0$