What is the minimum value of the fraction $\frac{4x^4+4y^4-3xy+5}{2x^2+2y^2-xy}$

Question

What is the minimum value of the fraction below ? $$\frac {4x^4+4y^4-3xy+5}{2x^2+2y^2-xy}$$ $$\tag {10 points }$$

This is a contest problem, but of course not an ongoing contest .

Our teacher solved the question with calculus methods. However, this was not the solution we were looking for.Because it's a high school competition.

My approach. I hope some inequalities help. However, they don't work very well. For example $$x^4+y^4\geq 2x^2y^2$$ or $$4x^4+4y^4\geq 8x^2y^2$$ Hence $$\frac{4x^4+4y^4-3xy+5}{2x ^2+2y^2-xy}\geq\frac{8x^2y^2-3xy+5}{2x^2+y^2-xy}$$ Also $$x^2+y^2\geq 2xy\Rightarrow 2x^2+2y^2\geq 4xy\Rightarrow 2x^2+2y^2-xy\geq 3xy$$ I don't know how AM-GM can help here.

I have a much more important note.

I really don't prefer fake solutions. What are fake solutions? Mathematically sound, methodically not. For example, we know what the minimum value of the fraction is beforehand, or we know beforehand under what conditions it takes this value and we manipulate the inequality applied accordingly. These methods add nothing to us, even if they are mathematically correct.

The solution I'm looking for is solutions that are instructive and not just "a lot of inequalities...and we are done".

Big thanks in advance.My purpose isn't just to get a solution, it's more than that.

Answer 1

Update.

Remark: Here is a solution without calculus (usually for high school problem).

We have $$f(x, y) := \frac{4x^4+4y^4-3xy+5}{2x^2+2y^2-xy} = 3 + \frac{4(x^4 + y^4) - 6(x^2 + y^2 ) + 5}{2(x^2 + y^2) - xy}.$$

Note that $4(x^4 + y^4) - 6(x^2 + y^2 ) + 5 = (4x^4 - 6x^2 + 5/2) + (4y^4 - 6y^2 + 5/2) > 0$ (easy).

We have \begin{align*} f &\ge 3 + \frac{4(x^4 + y^4) - 6(x^2 + y^2 ) + 5}{2(x^2 + y^2) + (x^2 + y^2)/2}\\[6pt] &\ge 3 + \frac{2(x^2 + y^2)^2 - 6(x^2 + y^2 ) + 5}{2(x^2 + y^2) + (x^2 + y^2)/2}\\[6pt] &= \frac35 + \frac45(x^2 + y^2) + \frac{2}{x^2 + y^2}\\[6pt] &\ge \frac35 + 2\sqrt{\frac45(x^2 + y^2) \cdot \frac{2}{x^2 + y^2}}\\[6pt] &= \frac35 + \frac45\sqrt{10} \tag{1} \end{align*} where we use $2(x^4 + y^4) \ge (x^2 + y^2)^2$ and $-xy \le (x^2 + y^2)/2$.

Also, when $x = -\frac12\sqrt[4]{10}, y = \frac12\sqrt[4]{10}$ (from the equality case of (1)), we have $f(x, y) = \frac35 + \frac45\sqrt{10}$.

Thus, the minimum is $\frac35 + \frac45\sqrt{10}$.

We are done.

Answer 2

Not a complete answer

Since both numerator and denominator are symmetric with respect to $y=x$ and $y=-x$ we can at first look for minimum along this paths to obtain:

1. $y=x$

$$\frac {4x^4+4y^4-3xy+5}{2x^2+2y^2-xy}=\frac83x^2+\frac53 \frac1{x^2}-1$$

with minimum $\frac{4\sqrt {10}}3-1$ at $y=x=\pm \sqrt[4]{\frac 58}$.

2. $y=-x$

$$\frac {4x^4+4y^4-3xy+5}{2x^2+2y^2-xy}=\frac85x^2+\frac1{x^2}+\frac35$$

with minimum $\frac{3+4\sqrt {10}}5$ at $y=-x=\pm \sqrt[4]{\frac 58}$.

The latter seems to be the minimum for the given ratio, but at the moment I can't justify this conclusion.

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We can try to generalize as follows, let $y=kx$ then

$$\frac {4x^4+4y^4-3xy+5}{2x^2+2y^2-xy}=\frac{(4k^4+4)x^4-3kx^2+5}{(2k^2-k+2)x^2}=$$

$$=\frac{4k^4+4}{2k^2-k+2}x^2+\frac{5}{2k^2-k+2}\frac1{x^2}-\frac{3k}{2k^2-k+2}=f(x)$$

with

$$f'(x) =\frac{8(k^4+1)x^4-10}{(2k^2-k+2)x^3}=0 \implies x^4=\frac{5}{4(k^4+1)}$$

and plugging into the original expression we reduce to

$$g(k)=\frac{4\sqrt 5\sqrt{k^4+1}-3k}{2k^2-k+2}$$

$$g'(k)=\frac{2(k^2-1)\left(3\sqrt{k^4+1}-2\sqrt 5 (k^2-4k+1)\right)}{(2k^2-k+2)^2\sqrt{k^4+1}}$$

which indicates $k=\pm 1$ as critical directions.