# Solve using AM GM Inequality if possible

Let x, y, z be non-zero real numbers such that $\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x} = 7$ and $\dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z} = 9$, then $\dfrac{x^3}{y^3} + \dfrac{y^3}{z^3} + \dfrac{z^3}{x^3}$ is equal to? I don't really know how to solve this. any methods would be welcome. I was solving a couple of AM GM inequality questions and I'm assuming this should be solved in a similar way. Correct me.

• for the use of AM-GM the variables must be nonnegative – Dr. Sonnhard Graubner Oct 24 '15 at 14:42
• Use Latex please, that will help to make the question readable. – Kushal Bhuyan Oct 24 '15 at 15:00
• First thought that comes to mind is use substitution $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$. – Cataline Oct 24 '15 at 15:33

Let $$\frac{x}{y} = a,\,\,\frac{y}{z} = b,\,\,\frac{z}{x} = c$$ where $$abc = \frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x} = 1$$

So we have $$\left\{ \begin{gathered} a + b + c = 7 \hfill \\ ab + ac + bc = 9 \hfill \\ \end{gathered} \right.$$

Using the identity $${a^3} + {b^3} + {c^3} - 3abc = (a + b + c)({a^2} + {b^2} + {c^2} - (ab + ac + bc))$$ which is equivalent to $${a^3} + {b^3} + {c^3} = {(a + b + c)^3} - 3(ab + ac + bc)(a + b + c) + 3abc$$ Now substituting the values we get $${a^3} + {b^3} + {c^3} = {7^3} - 3 \cdot 9 \cdot 7 + 3 = 343 - 189 + 3 = 157$$

Let $\displaystyle \frac{x}{y} =a \;\;,\frac{y}{z} = b\;\;,\frac{z}{x} = c\;,$ Then $a+b+c=7$ and $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 9$

and $abc=1$. Then we have to calculate $a^3+b^3+c^3 =$

Now Using $$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 9\Rightarrow \frac{ab+bc+ca}{abc} = 9\Rightarrow ab+bc+ca = 9$$

Now Let $t=a\;,t=b\;,t=c$ be the roots of cubic equation in terms of $t\;$ Then

Using factor Theorem, We get $(t-a)\;,(t-b)\;,(t-c)$ are the factor of that equation.

So we get $$(t-a)(t-b)(t-c) =0\Rightarrow t^3-(a+b+c)t^2+(ab+bc+ca)t-abc=0$$

So we get $$t^3-7t^2+9t-1=0\Rightarrow t^3=7t^2-9t+1$$

Now above we have take $t=a\;,t=b\;,t=c$ be the root of that cubic equation

So we get $$a^3+b^3+c^3=7(a^2+b^2+c^2)-9(a+b+c)+3 = 7(31)-9(7)+3=157$$

let $a= \frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, then some manipulations would show that

$(a+b+c)^3= 3(a+b+c)((a^2+b^2+c^2)) - 2(a^3+b^3+c^3) + 6abc$

we already have $a+b+c = 7, abc=1$, now its left to find what $a^2+b^2+c^2$ is.

$(\frac{x}{y} + \frac{y}{z}+ \frac{z}{x})^2 = (\frac{x}{y})^2 + (\frac{y}{z})^2+ (\frac{z}{x})^2 + 2(\frac{y}{x}+\frac{z}{y}+\frac{x}{z})$

$a^2+b^2+c^2=(\frac{x}{y})^2 + (\frac{y}{z})^2+ (\frac{z}{x})^2 = (\frac{x}{y} + \frac{y}{z}+ \frac{z}{x})^2 - 2(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}) = 7^2-2(9)=37$

so we have

$7^3 = 3(7)(37)-2(a^3+b^3+c^3)+6$

$((\frac{x}{y})^3+(\frac{y}{z})^3+(\frac{z}{x})^3) = 157$

This method doesn't use the $AM-GM$ inequality.

$a^3 +b^3 +c^3 = (a+b+c)^3 -3(a+b+c)(ab+ac+bc)+3abc$

Applying it to your question, we have

$\dfrac{x^3}{y^3} + \dfrac{y^3}{z^3} + \dfrac{z^3}{x^3}$

= $(\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x})^3 -3(\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x})(\dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z}) +3$

=$(7^3)-3(7)(9) +3$

=$157$