# Find the min value of $\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}$ when ${x^2} + {y^2} + {z^2} = 1$. [duplicate]

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.

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

Since

\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}}

$$a^2+b^2+c^2≥ab+bc+ac$$
$$(a+b+c)^2≥3(ab+bc+ac)\thinspace .$$
• 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 $(a+b+c)^2 = 3(ab+bc+ac)+\frac{1}{4}(a-2b+c)^2+\frac{3}{4}(a-c)^2$? Sep 17 at 16:14
• 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. :) Sep 17 at 18:50