If $x,y\in\mathbb{R}$ and $x^2+y^2>0.$ Then maximum and minimum value of $\displaystyle \frac{x^2+y^2}{x^2+xy+4y^2}$


Let $$K=\frac{x^2+y^2}{x^2+xy+4y^2}$$

$$Kx^2+Kxy+4Ky^2=x^2+y^2\Rightarrow (4K-1)y^2+Kxy+(K-1)x^2=0$$

put $y/x=t$ and equation is $(4K-1)t^2+Kt+(K-1)=0$

How do i solve it Help me plesse

  • $\begingroup$ And, just graphing it, only when $K$ is equal to $1$, does anything show up. When $K$ is equal to $1$, there are no minima or maxima. $\endgroup$
    – DUO Labs
    Jun 13, 2019 at 0:19
  • $\begingroup$ It is interesting that the function is like the reciprocal of the Rayleigh coefficient of a certain matrix, that would help. Indeed, the minimum and maximum of this function will be related with the eigenvalues of the mateix. Give me some time, I'll give it a try. $\endgroup$ Jun 13, 2019 at 0:56
  • $\begingroup$ @Quote Davis Try graphing $K$ as a function of $t$. $\endgroup$ Jun 13, 2019 at 3:55

3 Answers 3


$K=\frac{1+t^2}{1+t+4t^2}$ To find min and max, set $K'(t)=0$ or $t^2-6t-1=0$ and solve for $t$. Answer $t=3\pm \sqrt{10}$ leading to max and min $K=\frac{20\pm 6\sqrt{10}}{80\pm 25\sqrt{10}}$. or $1.088303688022443$ and $0.245029645310883$

Corrected (stupid error in solving quadratic).

  • $\begingroup$ How did you know to do this? Just insight or a traditional technique? $\endgroup$ Jun 13, 2019 at 2:43
  • $\begingroup$ @Chase Ryan Taylor standard calculus $\endgroup$ Jun 13, 2019 at 3:02
  • $\begingroup$ I meant the substitution $\endgroup$ Jun 13, 2019 at 3:41
  • $\begingroup$ I don't understand your question. Note the correction. $\endgroup$ Jun 13, 2019 at 3:46
  • $\begingroup$ What I mean is: For a general $f(x,y)$ to maximize, what is the formula for $K(t)$? $\endgroup$ Jun 13, 2019 at 4:17

Let $v = (x,y)$, $$ A = \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 4 \end{bmatrix} $$ and define the Rayleigh cuocient. $$ R_A(v) = \frac{v^T A v}{v^Tv}. $$ Then, it is clear that $$ \frac{x^2+y^2}{x^2+xy+4y^2} = \frac{1}{R_A(v)}. $$ It is well-known (Linear Algebra result) that $\operatorname{min}_{v \neq 0} R_A(v) = \lambda_{\operatorname{min}}$ and $\operatorname{max}_{v \neq 0} R_A(v) = \lambda_{\operatorname{max}}$, where $\lambda_{\operatorname{min}}$ and $\lambda_{\operatorname{max}}$ are the smallest and largest eigenvalues of $A$, respectively.

In this case $\lambda_{\operatorname{min}} = \frac{1}{2} \left(5-\sqrt{10}\right)$ and $\lambda_{\operatorname{max}} = \frac{1}{2} \left(5+\sqrt{10}\right)$. Therefore, $$ \min_{x,y\neq 0} \frac{x^2+y^2}{x^2+xy+4y^2} = \frac{1}{\max_{v\neq 0} R_A(v)} = \frac{1}{\frac{1}{2} \left(5+\sqrt{10}\right)} = \frac{2}{15}(5+\sqrt{10}) $$ and $$ \max_{x,y\neq 0} \frac{x^2+y^2}{x^2+xy+4y^2} = \frac{1}{\min_{v\neq 0} R_A(v)} = \frac{1}{\frac{1}{2} \left(5-\sqrt{10}\right)} = \frac{2}{15}(5-\sqrt{10}). $$

  • 1
    $\begingroup$ +1: Nice approach. $\endgroup$
    – copper.hat
    Jun 13, 2019 at 16:34

From the restriction $x^2+y^2>0$, we get that $x,y$ are not both zero.

If $x=0$, then $K={\large{\frac{1}{4}}}$.

Suppose $x\ne 0$.

Letting $t={\large{\frac{y}{x}}}$, and following your approach, we get $$(4K-1)t^2+Kt+(K-1)=0$$ which has a real solution for $t$ if and only $K={\large{\frac{1}{4}}}$ or the discriminant $$K^2-4(4K-1)(K-1)$$ is nonnegative.

Equivalently, either $K={\large{\frac{1}{4}}}$ or $$-15K^2+20K-4\ge 0$$ With the restriction $K\ne{\large{\frac{1}{4}}}$, the quadratic inequality solves as $$\frac{10-2\sqrt{10}}{15}\le K\le \frac{10+2\sqrt{10}}{15},\;\;K\ne{\large{\frac{1}{4}}}$$ Noting that $$\frac{10-2\sqrt{10}}{15}<\frac{1}{4}<\frac{10+2\sqrt{10}}{15}$$ it follows that

  • The minimum value of $K$ is ${\large{\frac{10-2\sqrt{10}}{15}}}\approx .2450296455$.$\\[8pt]$
  • The maximum value of $K$ is ${\large{\frac{10+2\sqrt{10}}{15}}}\approx 1.088303688$.

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