Minimize This!

If $\{ a,b\} \in \Re^{+}$ , then ${ (a-b) }^{ 2 }\ge 0$

$\Rightarrow{ a }^{ 2 }+{ b }^{ 2 }\ge 2ab$

$\Rightarrow \frac { { a }^{ 2 }+{ b }^{ 2 } }{ ab } \ge 2$

$\Rightarrow\frac { a }{ b } +\frac { b }{ a } \ge 2$

Thus, the minimum positive value of the given expression is $2$

simple.. the minimum positive number is 1.. putting both a and b 1 you get 1+1=2

Using the property of Arithmetic Mean>=Geometric Mean, (a/b+b/a)/2>=sqrt(a/b x b/a) (a/b+b/a)>=2

Let $\frac{a}{b} = x > 0$ then we wish to minimize $f(x) = x + \frac{1}{x} = \frac{x^2 + 1}{x} \rightarrow f '(x) = \frac{ 2x \cdot x - (x^2 + 1)}{x^2}=$ $\frac{x^2 - 1}{x^2} \rightarrow$ f '(x) = 0 if and only if x = 1 since x > 0, and f '(x) < 0 if $x \in (0,1)$ and f '(x) > 0 if $x \in (1, \infty)$ . Therefore, there is a minimum at x = 1 and f(1) =2 is the minimum of the expression above

After getting to $\dfrac{a^2+b^2}{ab}$ I considered that, to get the minimum value, since both of them are positive, we would need $a=b$ . Therefore $\dfrac{2a^2}{a^2}$ and the answer would be $2$ . Is this a valid solution?

Since $a,b\in\mathbb{R}^{+}$ , the minimum positive value of a,b is not $1$ . It's $0.0\ldots 01$

Using AM-GM, we will have $\frac{a}{b}+\frac{b}{a}\geq 2\sqrt{\frac{a}{b}\times \frac{b}{a}}=2$

Consider the triangle with lengths $a,~b>0 \mbox{ and }c=\sqrt{a^2 + b^2}$ . Let $\theta$ be the angle between $c$ and $a$ . Then, $\frac{a}{b}+\frac{b}{a} = \frac{c^2}{ab} = \frac{1}{\sin(\theta)\cos(\theta)} = \frac{2}{\sin(2\theta)}\leq 2$ . Hence, the result follows.

An easier Method will be by AM>GM :)

If a and b is positive then the least positive value for a and b is 1 Therefore, $\frac{a}{b}$ + $\frac{b}{a}$ = $\frac{1}{1}$ + $\frac{1}{1}$ =1+1 =2

Max-Min Inequality works here as well. If you set a=b, you will either get a max or a min. And by simply plugging in, one finds a min. You can also prove this if you are fluent with Multivariable Calculus.

By more thinking,the lowest +ve number is what +0.1 !!!! So by putting value of a & b by +0.1,then we find that solution is becomes 2. ;-) ;-) Technology!!!!! Technique....