Number $+$ Reciprocal

A simpler alternative proof:

$x + \frac {1}{x} \geq 2$

Since x > 0, we can multiply both sides by x :

$x^2 + 1 \geq 2x$

$x^2 - 2x + 1 \geq 0$

$(x - 1)^2 \geq 0$

Which is obviously true, because the square of a real number (x - 1) is always non-negative.

Actually there are many way to prove the given condition. Now I am going to prove it using AM and G.M inequality . It states that between any two numbers $a,b$ then the realation between AM (Arithmetic Mean) and GM (geometric Mean) between $a,b$ is that : $A.M \geq GM \implies \dfrac{a + b}{2} \geq (ab)^{1/2}$

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Coming to this question the given numbers are $x,\dfrac1x$ . Now,

$\begin{array}{c}~A.M \geq G.M \\ \underbrace{\dfrac{x + \dfrac1x}{2}}_{\color{#3D99F6}\text{A.M}} \geq \underbrace{(x \times \dfrac1x)^{1/2}}_{\color{#E81990}\text{G.M}} \\ \dfrac{x + \dfrac1x}{2} \geq 1 \\ (x + \dfrac1x) \geq 2 \\ \end{array}$ $\large \implies \boxed{\color{#20A900} \left(x + \dfrac1x \right)_{\text{min}} = 2} \quad \quad$

Let's assume there is a value of $x$ which results in a number less than $2$ :

$x+\frac{1}{x}<2$

We can rearrange this to:

$x^2-2x+1<0$

$(x-1)^2<0$

No real value of $x$ here can satisfy this equation. Therefore $x+\frac{1}{x}\ge 2$ .

$x+\frac{1}{x}=b$ $x^2-bx+1=0$ By the quadratic formula: $x=\frac{b+\sqrt{b^2-4}}{2}$ The positive domain of the function above is $\boxed{b≥2}$

I used the derivative:

$f(x)=x+\frac{1}{x}$

$f'(x)=1-\frac{1}{x^2}$

$f''(x)=\frac{2}{x^3}$

Let's see if we can find a minimum:

$f'(x)=0 \Rightarrow x=1$ since $x>0$

We have a maximum or minimum at $x=1$ . Let's check with $f''$ :

$f''(1)=2$

It's a minimum! So what's the minimum value of $f(x), x>0$ ?

$f(1)=1+1=2$

Definitely not the prettiest solution, but it's another way to do it.

For positive x, simple algebra gives

$(\sqrt{x}-\frac{1}{\sqrt{x}})^{2}=x+\frac{1}{x}-2$

Since the left hand side is the square of a real number it is non-negative, so

$x+\frac{1}{x}-2 \ge 0$

and the result follows!

Let's treat $x + \frac {1}{x}$ as a function. So we have $f(x) = x + \frac {1}{x}$

Then we want to find its minimum for positive values of $x$ . We can do this the usual way, by the first derivative.

$f'(x) = 1 - \frac {1}{x^2}$

$1 - \frac {1}{x^2} = 0$

$1 = \frac {1}{x^2}$

$x^2 = 1$

$x = 1$ or $x = -1$

But since $x$ has to be positive, then we choose $x = 1$

Let's verify that $x = 1$ is in fact a minimum with the second derivative.

$f''(x) = \frac {2}{x^3}$

For $x = 1$ we have $f''(1) = 2 > 0$

Therefore the value $x = 1$ give us a minimum.

Let's check that minimum for $f(x)$ when $x = 1$

$f(1) = 1 + \frac {1}{1} = 2$

So we can see that, not only the expression $x + \frac {1}{x}$ can be equal to two, but also that it is greater than 2 for any other positive value of $x$ .

$x + \frac {1}{x} \geq 2$ $\forall$ $x$ $\in$ $\mathbb R_{> 0}$

The smaller "x" gets, the bigger "1/x" gets, so an x value < 1 would never give the expression a value < 2, it reaches it's minimum value at x=1 and keeps increasing beyond that point.

just use basic inequality $a+b\geq 2\sqrt{ab}$ ( $a,b\geq 0$ )

note if $x = 1$ then $x + \frac {1}{x} = 2$

so write $x = 1+y, x>0$ So

$x+\frac {1}{x}$ $=(1+y)+\frac {1}{1+y}$

$= \frac {(1+y)^2+1}{1+y}$

$= \frac {y^2+2y+2}{1+y}$

$= \frac {y^2}{1+y} + \frac {2+2y}{1+y}$

$= \frac {y^2}{1+y} + 2$

$>= 2$

Note: this relies on $1+y > 0$ this as true as $1+y = x > 0$

For all x in R*+, ( x+1/x ) >= 2.

f(x)=x+1/x , If x=1, 1+1=2 , If x->0, you'll have (x+1/x) -> infinite , If x -> infinite,

lim x->0 (f(x))=infinite lim x->infinite (f(x))=infinite

f'(x)=1+ $\frac{-1}{x^2}$ =0 to find the minimum

$\frac{x^2-1}{x^2}$ =1+ $\frac{-1}{x^2}$ =0 so when x=1, this is the minimum, f(1)=2 So, f(x)>=2.

The graph to add more informations:

Let $x = 1+y$ where $y \geq 0$ and $x \geq 1$

Putting in main equation

$(1+y) + \frac {1}{1+y}$

$= \frac {(1+y)^2 + 1}{(1+y)}$

$= \frac {y^2+2y+2}{1+y }$

$= \frac {y^2}{y+1}+2 \geq 2$

And for

$x<1 \Rightarrow \frac1x >1$

so same equation will apply

The minimum positive number is 1. With it the value is 2. Hence for any other positive number it will always be greater than 2.

Not necessarily true, but it helps with the reasoning: let's assume that x is rational, x = $\frac{a}{b}$ . then x + $\frac{1}{x}$ = $\frac{x^2 + y^2}{xy}$ . Subtract 2 from both sides, for: x + $\frac{1}{x}$ - 2 = $\frac{x^2 - 2xy + y^2}{xy}$ = $\frac{(x-y)^2}{xy}$ . xy is positive, and $(x-y)^2$ is either 0, or positive, thus the original x + $\frac{1}{x}$ must be $\geq$ 2.

We have the function $f(x)=x+\frac {1}{x}$

Consider $f'(x)$ and $f''(x)$ :

$f'(x)=\frac {\mathrm d}{\mathrm dx} f(x)=1-\frac {1}{x^2}$

$f''(x)=\frac {\mathrm d}{\mathrm dx} f'(x)= \frac {2}{x^3}$

In order to find the minimum value of $f(x)$ , set the derivative to 0:

$0=1-\frac {1}{x^2}$

$\frac {1}{x^2}=1$

$x^2=1$

$x=\pm 1$

Since x is a positive number, $x=1$ . As $f''(1)=2>0$ , $f(x)$ has a minimum at $x=1$ .

$f(x) \geq f(1) = 2 \rightarrow \boxed {True}$

Let $x=1+\Delta x$ .

This allows the question to be rewritten: $(1+\Delta x) + \frac{1}{(1+\Delta x)} >= 2$

Combining terms and rearranging gives us: $\Delta x + \frac{1}{1+\Delta x} >= 1$

$\frac{\Delta x^2 + \Delta x + 1}{1+\Delta x} >= 1$

$\frac{\Delta x^2 }{1+\Delta x} + \frac{\Delta x + 1}{1+\Delta x} >= 1$

$\frac{\Delta x^2 }{1+\Delta x} >= 0$

$\Delta x^2 >= 0$

Which is true for any $\Delta x$

Probably the simplest solution is that A.M >- G.M. Then (x + 1/x)/2 >= sqrt((x)*(1/x) = sqrt(1) = 1. Multiplying by 2, x + 1/x >+ 2. A second solution is to set the first derivative = 0, then show this leads to a maximum. That is let f(x) = x + 1/x. Then f'(x) = 1 -I/(x^2) = 0, or x^2 = 1, and x >0 implies x = 1; then f(1) =2, and f''(x) <0 shows this is a maximum. A third solution is to solve the equality x + 1/x =2, or x^2 - 2x + 1 =0, x _1 )^2 = 0, x =1. Then substitute values just above and below 1, and we find The value of x + 1/x exceeds 2. Ed Gray

x+(1/x)=(x^2+1)/x=((x-1)^2+2x))/x=(x-1)^2/x+2 (x-1)^2/x>0 such that x+(1/x)=((x-1)^2/x)+2>2

The most basic solution none of this garbage y'all are posting about the arithmetic mean

( $x$ - 1) $^2$ > 0

$x^2$ +1 $>$ $2x$

$x$ + $\frac{1}{x}$ > 2

x + 1/x => 2
Multiply both sides of the inequality by x. Because x is a positive number, the inequality does not flip.
x^2 + 1 => 2x
x^2 -2x + 1 => 0
(x-1)^2 => 0
Because this final expression is squared, it cannot be negative. It may equal any non negative as well. Consider any a=>0:
x^2 -2x + 1 = a
x^2 -2x + (1-a) = 0
Apply the quadratic formula:
x = [ -(-2) +/- sqrt(2^2 - 4(1-a)(1)) ] / (2(1))
If the discriminant is non-negative, then the expression can equal a:
d = sqrt(2^2 - 4(1-a)(1))
d = sqrt(4 - 4 + a)
d = sqrt(a) => 0, since a=>0

Hence, x^2 - 2x + 1 = (x-1)^2 = a is possible for any a=>0, and the original equality is always true.

if n + 1/n -2 < 0 , then n² - 2n + 2 < 0, (n - 1)² + 1 < 0 (n - 1)² < -1 so it does not support

I don’t like the solutions involving (x - 1) squared. By that logic, x + 1/x >= 2 is true for negative numbers as well.

Here’s a different approach. We can observe that it is true for any numbers we try. For example, x = 9/10. 9/10 is less than 1 by 1/10, but 10/9 is greater than 1 by 1/9, so 9/10 + 10/9 > 2.

Let e>0 be an arbitrary small number. Let x be 1 - e. Then x + 1/x = 1 - e + 1/(1 - e) = 1 - e + (1 + e)/(1 - e^2). Since 1 - e^2 < 1, (1 + e)/(1 - e^2) = 1 + e + c where c is positive. Thus 1 - e + 1/(1 - e) = 2 + c > 2. Done.

Arithmatic mean>= Geometricmean so it's true

srx means square root of x.

Ok.

(srx - 1/srx)^2 >= 0 is always true

x - 2 + 1/x >= 0 Now add 2 on either side.

x + 1/x >= 2 is still always true.

There are 2 ways to prove:

1. AM >= GM Inequality

arethemetic mean is greater than or equal to geometric mean

=> AM of x And 1/x = (x +1/x)/2

=> GM of x and 1/ x = ( x*1/x) = 1

=> now AM >= GM So (x +1/x)/2 >= 1 or x +1/x >= 2

;2. Square of a real no. Is always >= 0

=> ( sqrt (x ) - sqrt (1/x))^2 >= 0

=> x + 1/x - 2 >= 0

=> x + 1/x >= 2