Solve it either Geometrically or Algebraically

Method-1 :

$a_{k} = x , b_{k} = y$

$x = lsina , y = lcosa$

$E = \dfrac{\dfrac{x}{l}.\dfrac{y}{l}}{ (\dfrac{x}{l})^{2} + (\dfrac{y}{l})^{2} + ( \dfrac{x}{l} + \dfrac{y}{l})\sqrt{(\dfrac{x}{l})^{2} + (\dfrac{y}{l})^{2}}}$

$E = \dfrac{sinacosa}{1 + (sina + cosa)\sqrt{1}}$

$2E = \dfrac{sin2a}{1 + sina + cosa}$

$2E = \dfrac{sin2a}{1 + \sqrt{2}sin(\dfrac{\pi}{4} + a)}$

for maximum denominator should be minimum

$(\dfrac{sin2a}{1 + \sqrt{2}sin(\dfrac{\pi}{4} + a)})_{max} = \sqrt{2} -1$ ( $a = 2n\pi + 2 tan^{-1}(\sqrt{2} -1)$

$E_{max} = \dfrac{\sqrt{2} - 1}{2}$

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Method-2:

$\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}}\ge\frac{x_1+\cdots+x_n}{n}$

$E = \dfrac{\dfrac{xy}{\sqrt{2}}}{\dfrac{x^{2} + y^{2}}{\sqrt{2}} + \dfrac{\sqrt{(x^{2} + y^{2})^{2}}}{2}}$ as we want minimum

$E = \dfrac{\sqrt{2}xy}{x^{2} + y^{2}(\sqrt{2} + 1)}$

$x^{2} + y^{2} \geq 2xy$

$E_{max} = \dfrac{1}{2(\sqrt{2} + 1)} = \dfrac{\sqrt{2} - 1}{2}$

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Method-3

Simply by $A.M \geq G.M$

$E = \dfrac{1}{\dfrac{x}{y} + \dfrac{y}{x} + \dfrac{(x + y)(\sqrt{x^{2} + y^{2}})}{xy}}$

$E_{max} = \dfrac{1}{ 2 + \dfrac{2\sqrt{xy}(\sqrt{2xy})}{xy}}$

$E_{max} = \dfrac{1}{2(\sqrt{2} + 1)} = \dfrac{\sqrt{2} - 1}{2}$

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This one's too easy

I have done it algebrically take ${a}_{k}+{b}_{k}=x$ and the whole problem reduces into finding the maximum value of :

$\large f(t)=E=\frac{x-l}{2l}$

Now using the standard equality(actually I forgot it's name) we get

$\large \frac{{a}_{k}^{2}+{b}_{k}^{2}}{2} \ge {(\frac{{a}_{k}+{b}_{k}}{2})}^{2}$

$\large x \le l\sqrt{2}$

Using this ${E}_{max}$ is :

$\large E= \frac{\sqrt{2}-1}{2}$

Just wondering Is there any geometrical solution to this?