## Friday, May 22, 2015

### Second Heuristic Method In Solving Problem: Compare $M$ and $N$.

Given that $p,\,q,\,r,\,s,\,a,\,b$ are positive real numbers with $M=\sqrt{ap+br}\cdot \sqrt{\dfrac{q}{a}+\dfrac{s}{b}}$ and $N=\sqrt{pq}+\sqrt{rs}$. Compare $M$ and $N$.

One of the other two inspiring methods of solving the above question will be discussed in this post.

When we're asked to compare two quantities, like in this instance, we have $M$ and $N$. But that doesn't restrict us in working and considering only with the linear power of  $M$ and $N$.

We could also consider the pairs like:

1.  [MATH]\color{yellow}\bbox[5px,purple]{M^2\,\,\text{and}\,\,N^2}[/MATH],

2.  [MATH]\color{yellow}\bbox[5px,blue]{M^3\,\,\text{and}\,\,N^3}[/MATH],

3.  [MATH]\color{yellow}\bbox[5px,orange]{M^4\,\,\text{and}\,\,N^4}[/MATH], and so on and so forth.

## Therefore, please DO NOT limit your thought and option to what are given. Think out of the box to bring a new perspective in.

I have seen one Taiwan's friend of mine who tackled this problem by comparing the quantities $M^2$ and $N^2$:

He first noticed

\begin{align*}N^2&=(\sqrt{pq}+\sqrt{rs})^2\\&=pq+rs+2\sqrt{pqrs}\end{align*}

\begin{align*}M^2&=\left(\sqrt{ap+br}\cdot \sqrt{\dfrac{q}{a}+\dfrac{s}{b}}\right)^2\\&=(ap+br)\left(\dfrac{q}{a}+\dfrac{s}{b}\right)\\&=ap\left(\dfrac{q}{a}\right)+ap\left(\dfrac{s}{b}\right)+br\left(\dfrac{q}{a}\right)+br\left(\dfrac{s}{b}\right)\\&=\cancel{a}p\left(\dfrac{q}{\cancel{a}}\right)+ap\left(\dfrac{s}{b}\right)+br\left(\dfrac{q}{a}\right)+\cancel{b}r\left(\dfrac{s}{\cancel{b}}\right)\\&=pq+rs+\dfrac{aps}{b}+\dfrac{brq}{a}---(*)\end{align*}

He then realized he could apply the AM-GM inequality to the last two terms in (*) to get:

\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*}

By wrapping this into (*) to obtain:

\begin{align*}M^2&= pq+rs+\dfrac{aps}{b}+\dfrac{brq}{a}\\&\ge pq+rs+2\sqrt{pqrs}\\&\ge N^2 \end{align*}

Taking square root of both sides gives us back

$M\ge N$

The other method compares the quantities $M^4$ and $N^4$, do you think you want to try that first before reading it?

You're encouraged to make an attempt to it, that is the only way you could learn with yourself!