Question

Given $x, y \in R, x^2 + y^2 > 0$. Then the range of $\frac{x^2+y^2}{x^2+xy+4y^2}$ is

Answer 1

$\left[\frac{10 - 2\sqrt{10}}{15}, \frac{10 + 2\sqrt{10}}{15}\right]$

Answer 2

Let the given expression be $E = \frac{x^2+y^2}{x^2+xy+4y^2}$.
We are given $x, y \in R$ and $x^2 + y^2 > 0$, which means $(x, y) \neq (0, 0)$.
To find the range of $E$, let $E = k$.
$k = \frac{x^2+y^2}{x^2+xy+4y^2}$
$k(x^2+xy+4y^2) = x^2+y^2$
$kx^2 + kxy + 4ky^2 = x^2 + y^2$
$(k-1)x^2 + kxy + (4k-1)y^2 = 0$.

This is a homogeneous equation of degree 2 in $x$ and $y$.

Case 1: $y = 0$.
Since $x^2+y^2 > 0$, if $y=0$, then $x^2 > 0$, so $x \neq 0$.
The equation becomes $(k-1)x^2 + kx(0) + (4k-1)(0)^2 = 0$, which simplifies to $(k-1)x^2 = 0$.
Since $x \neq 0$, we must have $k-1 = 0$, so $k=1$.
If $k=1$, the original expression is $\frac{x^2+y^2}{x^2+xy+4y^2} = \frac{x^2}{x^2} = 1$ (for $y=0, x \neq 0$).
So $k=1$ is in the range.

Case 2: $y \neq 0$.
We can divide the equation $(k-1)x^2 + kxy + (4k-1)y^2 = 0$ by $y^2$ to get a quadratic equation in $\frac{x}{y}$:
$(k-1)\left(\frac{x}{y}\right)^2 + k\left(\frac{x}{y}\right) + (4k-1) = 0$.
Let $t = \frac{x}{y}$. The equation is $(k-1)t^2 + kt + (4k-1) = 0$.

For real solutions for $t$ (which correspond to real solutions for $x$ and $y$ with $y \neq 0$), the discriminant of this quadratic equation must be non-negative, provided the equation is indeed quadratic (i.e., $k-1 \neq 0$).

If $k-1 \neq 0$, the discriminant is $\Delta = b^2 - 4ac = k^2 - 4(k-1)(4k-1)$.
$\Delta = k^2 - 4(4k^2 - k - 4k + 1) = k^2 - 4(4k^2 - 5k + 1) = k^2 - 16k^2 + 20k - 4 = -15k^2 + 20k - 4$.
For real solutions for $t$, we require $\Delta \ge 0$:
$-15k^2 + 20k - 4 \ge 0$
$15k^2 - 20k + 4 \le 0$.

To find the values of $k$ satisfying this inequality, we find the roots of the quadratic equation $15k^2 - 20k + 4 = 0$.
Using the quadratic formula $k = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$k = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(15)(4)}}{2(15)} = \frac{20 \pm \sqrt{400 - 240}}{30} = \frac{20 \pm \sqrt{160}}{30} = \frac{20 \pm 4\sqrt{10}}{30} = \frac{10 \pm 2\sqrt{10}}{15}$.
Let $k_1 = \frac{10 - 2\sqrt{10}}{15}$ and $k_2 = \frac{10 + 2\sqrt{10}}{15}$.
Since the quadratic $15k^2 - 20k + 4$ opens upwards, $15k^2 - 20k + 4 \le 0$ when $k$ is between the roots (inclusive).
So, $k_1 \le k \le k_2$, i.e., $\frac{10 - 2\sqrt{10}}{15} \le k \le \frac{10 + 2\sqrt{10}}{15}$.

This interval gives the possible values of $k$ for which the quadratic in $t$ has real solutions, provided $k \neq 1$.

If $k-1 = 0$, i.e., $k=1$, the equation $(k-1)t^2 + kt + (4k-1) = 0$ becomes $0 \cdot t^2 + 1 \cdot t + (4 \cdot 1 - 1) = 0$, which is $t + 3 = 0$.
This linear equation has a real solution $t = -3$. This corresponds to $\frac{x}{y} = -3$, or $x = -3y$.
For any $y \neq 0$, we can find $x = -3y$. Then $x^2+y^2 = (-3y)^2+y^2 = 9y^2+y^2 = 10y^2$. Since $y \neq 0$, $10y^2 > 0$, so $x^2+y^2 > 0$.
The value of the expression for $x=-3y$ is $\frac{(-3y)^2+y^2}{(-3y)^2+(-3y)y+4y^2} = \frac{9y^2+y^2}{9y^2-3y^2+4y^2} = \frac{10y^2}{10y^2} = 1$.
So $k=1$ is in the range.

We need to check if $k=1$ is included in the interval $\left[\frac{10 - 2\sqrt{10}}{15}, \frac{10 + 2\sqrt{10}}{15}\right]$.
$k_1 = \frac{10 - 2\sqrt{10}}{15}$. Since $2\sqrt{10} = \sqrt{40}$, and $6 < \sqrt{40} < 7$, $2\sqrt{10} \approx 6.32$.
$k_1 \approx \frac{10 - 6.32}{15} = \frac{3.68}{15} > 0$.
$k_2 = \frac{10 + 2\sqrt{10}}{15} \approx \frac{10 + 6.32}{15} = \frac{16.32}{15} > 1$.
Let's formally check $1$ against the bounds:
Is $1 \ge \frac{10 - 2\sqrt{10}}{15}$? $15 \ge 10 - 2\sqrt{10} \iff 5 \ge -2\sqrt{10}$, which is true.
Is $1 \le \frac{10 + 2\sqrt{10}}{15}$? $15 \le 10 + 2\sqrt{10} \iff 5 \le 2\sqrt{10}$. Squaring both sides (both are positive), $25 \le 40$, which is true.
So $k=1$ is indeed within the interval $\left[\frac{10 - 2\sqrt{10}}{15}, \frac{10 + 2\sqrt{10}}{15}\right]$.

The denominator $x^2+xy+4y^2$ cannot be zero for any $(x,y) \neq (0,0)$ such that $(k-1)x^2 + kxy + (4k-1)y^2 = 0$.
If $x^2+xy+4y^2 = 0$, then $\frac{x^2+y^2}{x^2+xy+4y^2}$ is undefined.
If $x^2+xy+4y^2 = 0$, then dividing by $y^2$ (if $y \neq 0$), $(x/y)^2 + (x/y) + 4 = 0$. Let $t=x/y$. $t^2+t+4=0$. The discriminant is $1^2 - 4(1)(4) = 1-16 = -15 < 0$. So $t^2+t+4=0$ has no real solutions for $t$. This means $x^2+xy+4y^2$ can never be zero if $y \neq 0$.
If $y=0$, $x^2+xy+4y^2 = x^2$. Since $x^2+y^2 > 0$, if $y=0$, then $x \neq 0$, so $x^2 \neq 0$.
Therefore, the denominator $x^2+xy+4y^2$ is always non-zero for $(x,y) \neq (0,0)$.
The derivation based on the quadratic in $t=x/y$ is valid for all $y \neq 0$. The case $y=0$ gives $k=1$, which is included in the interval obtained from the discriminant.

The range of the expression is the interval $\left[\frac{10 - 2\sqrt{10}}{15}, \frac{10 + 2\sqrt{10}}{15}\right]$.