Does the trick have something to do with the equation of a circle?
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$x=\cos t,y=\sin t$ – lab bhattacharjee Mar 21 '19 at 04:22
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Hint. Let $x=\sin\theta, y=\cos\theta$ to have $x^2+4xy-y^2=\sin^2\theta+4\sin\theta\cos\theta-\cos^2\theta$. Can you use calculus to minimise this function of one variable?
YiFan Tey
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1Why calculus $$2(\sin2\theta)-\cos2\theta\le\sqrt{2^2+(-1)^2}$$ – lab bhattacharjee Mar 21 '19 at 04:26
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That works as well. But if one didn't see that expression, calculus is the surefire way to go. – YiFan Tey Mar 21 '19 at 04:30
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Let $x^2+4xy-y^2=c$ and $\dfrac c1=\dfrac{x^2+4xy-y^2}{x^2+y^2}$
$\iff(c-1)(x/y)^2-4(x/y)+c+1=0$
As $x/y$ is real,the discriminant must be $\ge0$
$$4^2\ge4(c^2-1)\iff c^2\le5$$
The equality will occur if $\dfrac xy=\dfrac4{2(c-1)}$
Alternatively,
Let $x^2+4xy-y^2=u\implies x^2-y^2=u-4xy$
$1^2=(x^2-y^2)^2+(2xy)^2=(u-4xy)^2+(2xy)^2\iff20(xy)^2-8u(xy)+u^2-1=0$
The discriminant $(8u)^2-80(u^2-1)=16(5-u^2)$
lab bhattacharjee
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