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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.

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    $\begingroup$ 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 $x$ ;) $\endgroup$ – Stan Tendijck Feb 21 '19 at 0:39
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    $\begingroup$ If you want a purely algebraic proof, then notice that this is a consequence of $(\sqrt x - 1/\sqrt x)^2 \ge 0$. $\endgroup$ – user296602 Feb 21 '19 at 0:40
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    $\begingroup$ You have made a sign error in your second derivative (try and double check this). $\endgroup$ – Minus One-Twelfth Feb 21 '19 at 0:44
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    $\begingroup$ You could use that $(x-1)^2 \ge 0$ to show it. $\endgroup$ – randomgirl Feb 21 '19 at 1:05
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    $\begingroup$ 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. $\endgroup$ – Taladris Feb 21 '19 at 1:55
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$f(x) = x + x^{-1}, \; \Bbb 0 < x \in \Bbb R; \tag 1$

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

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

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

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

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

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

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

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

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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.

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