The third inspiring methods of solving the above question will be discussed in this post.
It's solved by one Indian genius, a math friend of mine who compared the quantities between $M^4$ and $N^4$:
He first noticed
$\begin{align*}N^4&=(\sqrt{pq}+\sqrt{rs})^4\\&=(pq+rs+2\sqrt{pqrs})^2---(1)\end{align*}$
He then stopped expanding. He planned, he did not rush to finish, he had hope to see if there are something worth to notice before he continued to expand blindly for the equation (1) above.
$\begin{align*}M^4&=(M^2)^2\\&=\left(\sqrt{(ap+br)\left(\dfrac{q}{a}+\dfrac{s}{b}\right)}^2\right)^2\\&=\left((ap+br)\left(\dfrac{q}{a}+\dfrac{s}{b}\right)\right)^2\\&=\left(pq+\dfrac{aps}{b}+\dfrac{brq}{a}+rs\right)^2\\&=\left(pq++rs+\dfrac{ap}{s}+\dfrac{brq}{a}\right)^2\end{align*}$
He stopped momentarily to compare at this point for the current expressions for both $M^4$ and $N^4$ and he realized one important thing at an instant:
[MATH]\color{black}N^4=(pq+rs+\color{yellow}\bbox[5px,purple]{2\sqrt{pqrs}}\color{black})^2[/MATH]
[MATH]\color{black}M^4=\left(pq++rs+\color{yellow}\bbox[5px,green]{\dfrac{aps}{b}+\dfrac{brq}{a}}\color{black}\right)^2[/MATH]
He knew if he could prove
[MATH]\color{yellow}\bbox[5px,green]{\dfrac{aps}{b}+\dfrac{brq}{a}}\color{black}\color{black}{\ge}\color{yellow}\bbox[5px,purple]{2\sqrt{pqrs}}[/MATH]
then he could conclude that $M\ge N$.
AM-GM inequality is the perfect weapon to use for this instance as we have:
$\begin{align*}\dfrac{aps}{b}+\dfrac{brq}{a}&\ge 2\sqrt{\dfrac{aps}{b}\cdot \dfrac{brq}{a}}\\&\ge 2\sqrt{\dfrac{\cancel{a}ps}{\cancel{b}}\cdot \dfrac{\cancel{b}rq}{\cancel{a}}}\\&\ge 2\sqrt{pqrs} \end{align*}$
Therefore, he concluded by now that $M\ge N$ for all positive real $p,\,q,\,r,\,s,\,a,\,b$.
The whole point of this and the previous second method of solving this problem is to get you learned something by watching the genuine awesome works of others, and always remember that:
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