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If a,b and c are positive real numbers then find the minimum value of $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$

Logically,a=b=c=1 but mathematically, I have tried to use some AM-GM, GM-HM inequalities but I am unable to solve this.

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  • $\begingroup$ Another approach: the function $f(a,b,c)=a/b+b/c+c/a$ has one critical point (where the partial derivatives vanish) where $a,b,c>0$. $\endgroup$ Commented Nov 10, 2016 at 17:40

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This is easy using the AM-GM inequality. We get $$ \frac{\frac ab + \frac bc + \frac ca}{3}\geq \sqrt[3]{\frac ab\cdot \frac bc \cdot \frac ca} = 1 $$

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  • $\begingroup$ Note that this only provides a bound; you also need to show that this bound can be achieved, e.g. with $a=b=c=1$. $\endgroup$
    – vadim123
    Commented Nov 10, 2016 at 17:41
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    $\begingroup$ @vadim123 The OP already showed that you can achieve the sum $3$ in his question. But yes, technically you are correct. $\endgroup$
    – Arthur
    Commented Nov 10, 2016 at 17:42
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    $\begingroup$ youtube.com/watch?v=hou0lU8WMgo $\endgroup$
    – vadim123
    Commented Nov 10, 2016 at 17:43
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Another approach is to use rearrangement inequality.
$\dfrac ab+\dfrac bc+\dfrac ca\geq\dfrac aa+\dfrac bb+\dfrac cc=3$

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  • $\begingroup$ Similarly, this is only a bound; one needs to show that it can be achieved to prove that $3$ is the minimum. $\endgroup$
    – vadim123
    Commented Nov 10, 2016 at 17:42
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    $\begingroup$ @vadim123 Since OP has already mentioned the case a=b=c=1, I am not going to mention it again. But yes, the minimum is attainable when a=b=c=1. $\endgroup$ Commented Nov 10, 2016 at 17:43
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I thought it might be instructive to present a way forward that forgoes appeal to well-know inequalities, but rather relies on straightforward use of calculus. To that end, we proceed.

Let $x=\frac ab$ and $y=\frac bc$. Then, we can write

$$\frac ab+\frac bc+\frac ca=x+y+\frac1{xy}$$

Now let $f(x,y)=x+y+\frac1{xy}$. We see that $$\begin{align}f_1(x,y)&=1-\frac1{yx^2}=0\implies x^2=y \tag 1\\\\ f_2(x,y)&=1-\frac1{xy^2}=0\implies y^2=x \tag 2 \end{align}$$

Putting $(1)$ and $(2)$ together reveals if $f_1(x,y)=f_2(x,y)=0$, then $x=y=1$.

Now, since $f_{11}(x,y)f_{22}(x,y)-f_{12}^2(x,y)=\frac{3}{(xy)^4}>0$, $f$ attains a local minimum value when $x=y=1$. Restricting $x>0$ and $y>0$, we see that for positive $x$ and $y$, the local minimum is a global one.

Finally, when $x=y=1$, $a=b=c=1$ and we have

$$\min_{(a,b,c)}\left(\frac ab+\frac bc+\frac ca\right)=1$$

for $a>0$, $b>0$, and $c>0$.

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AM-GM inequality works : let's call $x_1=\frac{a}{b}$, $x_1=\frac{b}{c}$ and $x_3=\frac{c}{a}$. We have $$\frac{x_1+x_2+x_3}{3}\ge\sqrt[3]{x_1x_2x_3}=1$$ Therefore the minimal value is greater than $3$, and $3$ can be obtained if (and only if) $a=b=c$.

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Without loss of generality, assume that: $c =min\{a\,\,b\,\,c\}$

$\frac{a}{b} +\frac{b}{c} +\frac{c}{a}$

=$\frac{(a-b)^2}{ab} +\frac{(a-c)(b-c)}{ac} +3 \geq 3$

Or another way! (I will post it soon, now I'm busy)

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Another way!

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{c^2 (a-b)^2+ab(b-c)^2 +bc(a-c)^2}{abc(b+c)} +3 \geq 3$

Equality holds when $a=b=c$

P/s: I found this identity a long time ago, I don't remember how I found it. Can you help me?

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