I don't have problems usually with these types of inequalities. They usually have a simple trick where you either take them two by two or else to get the question. Or factor out $x + y + z$ etc.

However when I tried to solve this it became kinda challenging.

I proved through AM-GM that : $$\frac{xy}{z} + \frac{yz}{x} + \frac{xz}{y} ≥ x + y + z $$

But this doesn't really help , as the condition is with ${x^2} + {y^2} + {z^2} = 1$.

I tried some other methods , but I failed. Any help is welcomed.


1 Answer 1


We observe that, if $x,y,z>0$ does not hold, then the global minimum does not exist . Therefore, we need the restriction $x,y,z>0$ .

Let $\thinspace\dfrac {xy}{z}=a,\thinspace \dfrac {xz}{y}=b,\thinspace \dfrac {yz}{x}=c$ . Then, the original problem is equivalent to :

$$\min \{a+b+c\}=X$$

where $ab+bc+ac=1$ and $a,b,c>0$ .


$$ \begin{align}(a+b+c)^2&≥3(ab+bc+ac)=3\end{align}\tag 1 $$

We conclude that :

$$ \begin{align}|a+b+c|=a+b+c≥\sqrt 3\thinspace\thinspace\thinspace\thinspace\thinspace\tiny{\blacksquare}\end{align} $$

Indeed, since we need the condition $a+b+c>0$ here, this forces the condition $x,y,z>0$ to be satisfied .

$(1)\thinspace\thinspace\thinspace $ Note that, the inequality

$$ \small {\begin{align}\frac {(a-b)^2}{2}+\frac {(b-c)^2}{2}+\frac {(c-a)^2}{2}≥0\end{align}} $$

leads to :


which is equivalent to :

$$(a+b+c)^2≥3(ab+bc+ac)\thinspace .$$

  • $\begingroup$ Where did you get the $(a+b+c)^2\geq 3(ab+bc+ac)$ from? Is it some well known theorem or follows immediately from one? Or do we have to use the principle axis theorem and show $(a+b+c)^2 = 3(ab+bc+ac)+\frac{1}{4}(a-2b+c)^2+\frac{3}{4}(a-c)^2$? $\endgroup$ Sep 17 at 16:14
  • $\begingroup$ @ReinhardMeier Please, give me some time and I will expand the answer. Thank you for your feedback . $\endgroup$ Sep 17 at 16:16
  • $\begingroup$ @ReinhardMeier I expanded the answer as much as I could . $\endgroup$ Sep 17 at 16:36
  • $\begingroup$ Yes. Nice solution. Would it make sense to mention that the minimum is attained for $$ x=y=z=a=b=c=\frac{1}{\sqrt{3}}\,? $$ I only see lots of $\ge$ signs. :) $\endgroup$
    – Kurt G.
    Sep 17 at 18:50
  • $\begingroup$ @KurtG. This would definitely be better. Let me add. Thank you for your feedback . $\endgroup$ Sep 17 at 19:24

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