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

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.

• Actually you made a small error. $x=1$ is actually the minimum of $f$ (over the positive domain). Filling in, leads to $f(1)=0$ and thus $f(x)\geq0$ for all $x$ ;) – Stan Tendijck Feb 21 '19 at 0:39
• If you want a purely algebraic proof, then notice that this is a consequence of $(\sqrt x - 1/\sqrt x)^2 \ge 0$. – user296602 Feb 21 '19 at 0:40
• You have made a sign error in your second derivative (try and double check this). – Minus One-Twelfth Feb 21 '19 at 0:44
• You could use that $(x-1)^2 \ge 0$ to show it. – randomgirl Feb 21 '19 at 1:05
• More generally, if $a$ and $b$ are positive real numbers, then $\frac{a}{b}+\frac{b}{a}\geqslant 2$. The precalculus proofs apply with a slight modification, while the calculus proofs probably require multivariable calculus. – Taladris Feb 21 '19 at 1:55

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

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

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

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

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

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

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

$$F(x)= x+ \frac 1x -2 , x>0$$ 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.

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

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.