Question

Given x, y ∈ R, $x^2+y^2>0$. If the maximum and minimum value of the expression $E = \frac{x^2+y^2}{x^2+xy+4y^2}$ are M and m, and A denotes the average value of M and m, compute (2016)A.

Answer 1

1344

Answer 2

1. Homogeneous Expression: The given expression $E = \frac{x^2+y^2}{x^2+xy+4y^2}$ is homogeneous. This means we can divide the numerator and denominator by $y^2$ (assuming $y \ne 0$) to express $E$ as a function of a single variable $t = x/y$.

2. Form a Quadratic Equation: Substitute $t=x/y$ into the expression for $E$, which gives $E = \frac{t^2+1}{t^2+t+4}$. Rearrange this equation to form a quadratic equation in $t$: $(E-1)t^2 + Et + (4E-1) = 0$.

3. Use Discriminant Condition: For $t$ to be a real number, the discriminant ($D$) of this quadratic equation must be non-negative ($D \ge 0$). Calculate $D = E^2 - 4(E-1)(4E-1)$.

4. Solve Inequality for E: The discriminant condition leads to the inequality $15E^2 - 20E + 4 \le 0$. Solve for the roots of $15E^2 - 20E + 4 = 0$ using the quadratic formula, which are $E = \frac{10 \pm 2\sqrt{10}}{15}$.

5. Determine Max and Min Values: Since the parabola $15E^2 - 20E + 4$ opens upwards, the inequality $15E^2 - 20E + 4 \le 0$ holds for $E$ values between the roots. Thus, $m = \frac{10 - 2\sqrt{10}}{15}$ and $M = \frac{10 + 2\sqrt{10}}{15}$.

6. Check Edge Case ($y=0$): If $y=0$, $E = \frac{x^2}{x^2} = 1$. This value lies within the determined range, confirming the maximum and minimum values.

7. Calculate Average and Final Result: Compute the average $A = \frac{M+m}{2} = \frac{20/15}{2} = \frac{4/3}{2} = \frac{2}{3}$. Finally, calculate $(2016)A = 2016 \times \frac{2}{3} = 1344$.