Using AM-GM to find the minimum

Question

Find the minimum value of $\dfrac{7x^{2} - 2xy + 3y^{2}}{x^{2} - y^{2}}$ if $x$ and $y$ are positive real numbers such that $x > y$.

This is a question from the 22nd Philippine Mathematical Olympiad. I have no idea how to solve this. I am curious as to how it was solved. Can anyone help me, please?

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Update: I found a solution here and it looks like it is longer than expected.

Answer 1

Let, $x=a+b$ and $y=a-b$ where $a>b$, we have

$$f(x,y)= \dfrac {2a^2+3b^2}{ab}+2$$

$$\implies \underbrace{\dfrac { 2a^2+3b^2}{ab}≥\dfrac{2\sqrt 6 ab}{ab}}_{\text{ arithmetic-geometric mean inequality}}=2 \sqrt 6$$

$$\implies \text{min} [ f(x,y)]= 2+2 \sqrt 6$$

REMARK.

According to equality condition of arithmetic-geometric mean inequality, we have

$$2a^2=3b^2 \Longrightarrow a= \dfrac {b\sqrt 3}{\sqrt 2}$$

which follows

$$y := b \sqrt {\dfrac {3}{2}}-b>0$$

So, conditions are hold.

Answer 2

Hint: put $t= \dfrac{x}{y} \implies t > 1$. Then consider : $f(t) = \dfrac{7t^2 - 2t + 3}{t^2 - 1}, t > 1$. Can you take it from here using calculus tools ?

Answer 3

Without using calculus and arithmetic-geometric mean inequality:

Substitute $u=\dfrac xy$ and $u>1$ then,

$$\dfrac {7u^2-2u+3}{u^2-1}=a, a>0$$

$$\implies u^2(7-a)-2u+(3+a)=0$$

$$\implies \Delta =1-(7-a)(3+a)≥0, ~ \text {where} ~0<a≠7 $$

$$\implies a≥ 2+2\sqrt 6$$

$$\implies \text {min} [a]=2+2\sqrt 6<7$$

Finally,

$$u=\dfrac {1}{5-2\sqrt 6}>1$$ which implies $x>y.$

This means, conditions are hold.

Answer 4

$$\dfrac{7x^{2} - 2xy + 3y^{2}}{x^{2} - y^{2}} \implies \dfrac{4 x}{x - y} + \dfrac{6 x}{x + y} - 3$$

I'm not sure this will help you but, heuristically, the minimum for $x,y>0$ always occurs when $x\approx 10y \longrightarrow min\approx 6.898989899 \quad $ If you start with any $y\in\mathbb{N}$ and step down from $10y$, you will find the minimum $x$ to begin with or you will find a point where $f(10y-n,y)$ is greater than $f(10y-(n-1),y).\quad$ At this point, the latter is the minimum $f(x,y)$.