Show that $\forall x>0,\;\frac{1}{x}+x\geq 2$

Question

Using optimality conditions, we want to show that for all positive $x$, we have $$\frac{1}{x}+x\geq 2\quad (1).$$

I think that the above problem is equivalent to showing that $1/x+x-2\geq 0, \forall x>0$. Let $f(x)=1/x+x-2$. Then I used the optimality conditions to get: $$f'(x)=-\frac{1}{x^2}+1$$ which is equal to zero iff $x=1$. And also $$f''(x)=-\frac{2}{x^3}$$ which is positive iff $x$ is negative or negative iff $x$ is positive. So, for $x>0$ we have a maximum point at $x=1$. But can we show that $f$ is non-negative?

I am not sure what I did wrong here. I'd appreciate any help. Thank you.

Answer 1

$f(x) = x + x^{-1}, \; \Bbb 0 < x \in \Bbb R; \tag 1$

$f'(x) = 1 - x^{-2} = 0 \Longrightarrow x = 1; \tag 2$

$f''(x) = 2x^{-3}; \tag 3$

$f''(1) = 2 \Longrightarrow 1 \; \text{is a local minimum of} \; f(x); \tag 4$

note that

$0 < x < 1 \Longrightarrow f'(x) < 0; \; 1 < x \Longrightarrow f'(x) > 0, \tag 5$

which implies that in fact $x = 1$ is a global minimum for $f(x)$; also,

$f(1) = 2, \tag 6$

and thus we conclude that

$f(x) \ge 2, \; \forall 0 < x \in \Bbb R. \tag 7$

Answer 2

$$ F(x)= x+ \frac 1x -2 , x>0$$ We can write, $$ F(x)=(\sqrt x)^2 + (\frac{1}{\sqrt x})^2 - 2.\sqrt x.\frac{1}{\sqrt x},{\ }since {\ }x>0$$
$$F(x)= (\sqrt x -\frac{1}{\sqrt x})^2$$ Thus we can conclude, $F(x) \geq 0 \forall x > 0$

Equality holds for x = 1.

Answer 3

Or you use basics of inequalities :P $$\dfrac{(x-1)^2}{x} \geq 0 \Longleftrightarrow x>0$$

Answer 4

$(x-1)^2\ge 0\iff x^2-2x+1\ge 0\iff x^2+1\ge 2x\iff x+\dfrac 1x\ge 2$ whenever $x>0$.

Answer 5

This is easy to do with a proof by contradiction.

Given the initial problem (1)

$$ \forall x \in \mathbb{R}_{>0} . \frac{1}{x} + x \ge 2 \tag{1} $$

Let's try to prove that the negation (102) is false.

$$ \frac{1}{c} + c < 2 \;\;\;\; \text{where $c$ is a positive real constant} \tag{102} $$

$c$ is positive, so we can multiply both sides by $c$ without flipping the inequality.

$$ 1 - 2c + c^2 < 0 \tag{103} $$

Notice that the LHS is a perfect square

$$ (1-c)^2 < 0 \tag{104} $$

(104) means that the LHS is nonnegative always and therefore not less than zero. That's a contradiction, therefore (1) is true.