Easier than it looks :)

The inequality can be written as

$x + \dfrac{1}{x} + y + \dfrac{1}{y} \ge N$ ,

where $x = \dfrac{a_{1}}{a_{2}}$ and $y = \dfrac{a_{3}}{a_{4}}$ .

Now by the AM-GM inequality we know that $b + \dfrac{1}{b} \ge 2$ for any positive real number $b$ , so $N = 2 + 2 = \boxed{4}$ .

If we expand the denominator, we find that the LHS is equivalent to:

$\frac { { a }_{ 1 }{ a }_{ 2 }{ { a }_{ 3 } }^{ 2 }+{ a }_{ 1 }{ a }_{ 2 }{ { a }_{ 4 } }^{ 2 }+{ a }_{ 3 }{ a }_{ 4 }{ { a }_{ 1 } }^{ 2 }+{ a }_{ 3 }{ a }_{ 4 }{ { a }_{ 2 } }^{ 2 } }{ { a }_{ 1 }{ a }_{ 2 }{ { a }_{ 3 }{ a }_{ 4 } } }$

From AM-GM we can find that:

$\frac { { a }_{ 1 }{ a }_{ 2 }{ { a }_{ 3 } }^{ 2 }+{ a }_{ 1 }{ a }_{ 2 }{ { a }_{ 4 } }^{ 2 }+{ a }_{ 3 }{ a }_{ 4 }{ { a }_{ 1 } }^{ 2 }+{ a }_{ 3 }{ a }_{ 4 }{ { a }_{ 2 } }^{ 2 } }{ 4 }\ge \sqrt [ 4 ]{ { a }_{ 1 }{ a }_{ 2 }{ { a }_{ 3 } }^{ 2 }{ a }_{ 1 }{ a }_{ 2 }{ { a }_{ 4 } }^{ 2 }{ a }_{ 3 }{ a }_{ 4 }{ { a }_{ 1 } }^{ 2 }{ a }_{ 3 }{ a }_{ 4 }{ { a }_{ 2 } }^{ 2 } }$

Which is the same thing as:

$\frac { { a }_{ 1 }{ a }_{ 2 }{ { a }_{ 3 } }^{ 2 }+{ a }_{ 1 }{ a }_{ 2 }{ { a }_{ 4 } }^{ 2 }+{ a }_{ 3 }{ a }_{ 4 }{ { a }_{ 1 } }^{ 2 }+{ a }_{ 3 }{ a }_{ 4 }{ { a }_{ 2 } }^{ 2 } }{ 4 }\ge { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }{ { a }_{ 4 } }$

To answer our question we can convert this to:

$\frac { { a }_{ 1 }{ a }_{ 2 }{ { a }_{ 3 } }^{ 2 }+{ a }_{ 1 }{ a }_{ 2 }{ { a }_{ 4 } }^{ 2 }+{ a }_{ 3 }{ a }_{ 4 }{ { a }_{ 1 } }^{ 2 }+{ a }_{ 3 }{ a }_{ 4 }{ { a }_{ 2 } }^{ 2 } }{ { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }{ { a }_{ 4 } } } \ge 4$

Therefore, $N=\boxed { 4 }$

The expression on the left can be simplified as $\frac{a_{3}^{2}+a_{4}^{2}}{a_{3}a_{4}}+\frac{a_{1}^{2}+a_{2}^{2}}{a_{1}a_{2}}$ .

Cauchy-Schwarz tells us that for $x,y$ in $\mathbb{R},(x+y)^{2}\leq2(x^{2}+y^{2})$ , so $2xy\leq x^{2}+y^{2} \Rightarrow \frac{x^{2}+y^{2}}{xy}\geq2$ .

Thus, $\frac{a_{1}^{2}+a_{2}^{2}}{a_{1}a_{2}}\geq2$ , $\frac{a_{3}^{2}+a_{4}^{2}}{a_{3}a_{4}}\geq2 \Rightarrow \frac{a_{3}^{2}+a_{4}^{2}}{a_{3}a_{4}}+\frac{a_{1}^{2}+a_{2}^{2}}{a_{1}a_{2}}\geq\boxed{4}$ .

WLOG, we assume that ${a}_{1}={a}_{2}={a}_{3}={a}_{4}=x$ . Hence, we have,

$\frac{ x^{ 2 }(2x)+x^{ 2 }(2x) }{ x^{ 4 } } \geq N$ \ $\frac{4x(x^{2})}{x^{4}} \geq N$ \ $\frac{4x}{x^{2}} \geq N$ \ $\frac{4}{x} \geq N$ \

${x}$ is a positive integer. ${N}$ maximized when ${x=1}$ . Therefore, the maximum value of ${N=4}$ .

${ ({ a }_{ 1 }-{ a }_{ 2 }) }^{ 2 }\ge 0$

$\therefore \quad { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }>2{ a }_{ 1 }a_{ 2 }\quad \quad and\quad { a }_{ 3 }^{ 2 }+{ a }_{ 4 }^{ 2 }>2{ a }_{ 3 }a_{ 4 }$

Therefore $\frac { { { a }_{ 1 }{ a }_{ 2 }\left( { a }_{ 3 }^{ 2 }+{ a }_{ 4 }^{ 2 } \right) +{ a }_{ 3 }{ a }_{ 4 }\left( { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 } \right) } }{ { a }_{ 1 }a_{ 2 }{ a }_{ 3 }{ a }_{ 4 } } \ge \frac { { { a }_{ 1 }{ a }_{ 2 }\left( { 2a }_{ 3 }{ a }_{ 4 } \right) +{ a }_{ 3 }{ a }_{ 4 }\left( { 2a }_{ 1 }a_{ 2 } \right) } }{ { a }_{ 1 }a_{ 2 }{ a }_{ 3 }{ a }_{ 4 } }$

$\therefore \quad N\quad =\quad \boxed { 4 }$

re-writing the inequality gives us:

$\frac {a_1a_2(a_3^2+a_4^2)+a_3a_4(a_1^2+a_2^2)} {a_1a_2a_3a_4} = N \Leftrightarrow \frac {a_3} {a_4} + \frac {a_4} {a_3} + \frac {a_2} {a_1} + \frac {a_1} {a_2}=N$

And by $AM-GM$ :

$\frac {a_3} {a_4}+\frac {a_4} {a_3}+\frac {a_2} {a_1}+\frac {a_1} {a_2} = N \geq 4\sqrt[4]{\frac {a_3} {a_4} \cdot \frac {a_4} {a_3} \cdot \frac {a_2} {a_1} \cdot \frac {a_1} {a_2}}=4$

${ (x-1) }^{ 2 }=x^{ 2 }-2x+1=x-2+\frac { 1 }{ x } \ge 0\quad \rightarrow \quad x+\frac { 1 }{ x } \ge 2\quad \rightarrow \quad \frac { { a }_{ 1 } }{ { a }_{ 2 } } +\frac { { a }_{ 2 } }{ { a }_{ 1 } } \ge 2\quad and\quad \frac { { a }_{ 3 } }{ { a }_{ 4 } } +\frac { { a }_{ 4 } }{ { a }_{ 3 } } \ge 2\quad \rightarrow \quad N\ge 4$

Just make all values of a = 1

The inequality can be rewritten as

$\dfrac{a_{3}^{2} + a_{4}^{2}}{a_{3}a_{4}} + \dfrac{a_{1}^{2} + a_{2}^{2}}{a_{1}a_{2}} \geq N$

Observe that $(a_{3}+a_{4})^{2} = a_{3}^{2} + 2a_{3}a_{4} + a_{4}^{2}$

and

$(a_{1}+a_{2})^{2} = a_{1}^{2} + 2a_{1}a_{2} + a_{2}^{2}$

That can be rewritten as:

$a_{3}^{2} + a_{4}^{2} = (a_{3}+a_{4})^{2} - 2a_{3}a_{4}$

and

$a_{1}^{2} + a_{2}^{2} = (a_{1}+a_{2})^{2} - 2a_{1}a_{2}$

Substituting those values into the inequality

$\dfrac{(a_{3} + a_{4})^{2} - 2a_{3}a_{4}}{a_{3}a_{4}} + \dfrac{(a_{1}+1_{2})^{2} - 2a_{1}a_{2}}{a_{1}a_{2}} \geq N$

This simplifies to

$\dfrac{(a_{3} + a_{4})^2}{a_{3}a_{4}} + \dfrac{(a_{1} + a_{2})^2}{a_{1}a_{2}} \geq N + 4$

Thus, the answer is 4.

It doesn't say we can't do it, so why not make $a_{1}$ = $a_{2}$ = $a_{3}$ = $a_{4}$ . To make things simpler, let's write all these equal terms as a:

$\frac{aa(a^{2}+a^{2}) + aa(a^{2}+a^{2})}{aaaa}\geq$ N

The problem is asking for the maximum constant N - how far can we push N? After what value will the inequality stop being true?

So, what we are looking for is the value of N, which makes the expression an equality:

$2[a^{2}(a^{2}+a^{2})]$ = N $a^{4}$

$2[(a^{4}+a^{4})]$ = N $a^{4}$

$2[(2a^{4})]$ = N $a^{4}$

$4a^{4}$ = N $a^{4}$

The correct answer should be $\boxed{4}$