How to get the minimum and maximum value of algebraic expression?

Question

here is the link of the question What will be the minimum value of the expression

Answer 1

Instead of trying to go about the problem as a whole, break it up into smaller parts. Define $$\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2 \quad\text{as} \quad f(x)+g(y)$$

Where $$f(x) = \left(x+\frac{1}{x}\right)^2 \quad \text{and} \quad g(y) = \left(y+\frac{1}{y}\right)^2$$

You'll quickly notice that these functions are (basically) identical. Their derivatives are $$f'(x) = 2x-\frac{2}{x^3} \quad \text{and} \quad g'(y) = 2y-\frac{2}{y^3}$$ They have global minimums of $4$ at $1 \; \text{or}\;-1$. Unfortuantely for us, that does not add to $1$! So we dig a little deeper, we know that both $f(x)$ and $g(y)$ have asymptotes at $0$ and that they grow exponentially as they approach $0$. From this we can devise a strategy to find the minimum. The best way to do this is to just split $1$ right down the middle and have $x=0.5$ and have $y=0.5$ as well. This will return $12.5$ answer C.

The reasoning behind splitting one in half is that because the functions have asymptotes as they approach zero, shifting either function even slightly will cause an exponential increase. Even if we lower one value to $0.25$, the value of the other at $0.75$ outweighs it.

Answer 2

By the generalized means inequality:

$$ \sqrt{\frac{x^2+y^2}{2}} \ge \frac{x+y}{2} \ge \sqrt{\frac{2}{\cfrac{1}{x^2}+\cfrac{1}{y^2}}} \quad\implies\quad \begin{cases}x^2+y^2 \ge 2 \cdot \left(\dfrac{x+y}{2}\right)^2 = \dfrac{1}{2} \\[10px] \dfrac{1}{x^2}+\dfrac{1}{y^2} \;\ge\; \dfrac{8}{(x+y)^2}=8 \end{cases} $$

It follows that the following inequality holds, with equality iff $\,x=y=\dfrac{1}{2}\,$:

$$\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2 = x^2+y^2 + \frac{1}{x^2}+\frac{1}{y^2}+4 \ge \frac{1}{2}+8 + 4 = \dfrac{25}{2}$$