# An Inequality with Cyclic Sums on Both Sides II

### Solution 1

Introduce new variables, with $x^{90}=ab^2c^3,\,$ $y^{90}=bc^2a^3,\,$ $z^{90}=ca^2b^3,\,$ and $x,y,z\gt 0.\,$ We have $(xyz)^{90}=(abc)^6,\,$ i.e., $(xyz)^{15}=abc\,$ and $(xyz)^{120}=(abc)^8.\,$ The required inequality is equivalent to

$x^{15}+y^{15}+z^{15}\ge (x^3+y^3+z^3)(xyz)^4.$

By the AM-GM inequality,

\displaystyle \begin{align} 7x^{15}+4y^{15}+4z^{15}&\ge 15\sqrt[15]{(x^{15})^7(y^{15})^4(z^{15})4}=15x^7y^4z^4\\ 7y^{15}+4z^{15}+4x^{15}&\ge 15\sqrt[15]{(y^{15})^7(z^{15})^4(x^{15})4}=15y^7z^4x^4\\ 7z^{15}+4x^{15}+4y^{15}&\ge 15\sqrt[15]{(z^{15})^7(x^{15})^4(y^{15})4}=15z^7x^4y^4 \end{align}

\displaystyle \begin{align} 15(x^{15}+y^{15}+z^{15})&\ge 15(x^3+y^3+z^3)(xyz)^4\,\Leftrightarrow\\ x^{15}+y^{15}+z^{15}&\ge (x^3+y^3+z^3)(xyz)^4. \end{align}

### Solution 2

Since the inequality is homogeneous, we may assume $abc=1.\,$ Then the inequality becomes

$\displaystyle \sum_{cycl}\sqrt[6]{\frac{a}{b}}\ge\sqrt[30]{\frac{a}{b}}.$

By the AM-GM inequality,

\displaystyle\begin{align} 2\sqrt[6]{\frac{a}{b}}+\sqrt[6]{\frac{b}{c}}+\sqrt[6]{\frac{c}{a}}+1\ge 5\cdot\sqrt[30]{\frac{a}{b}}\\ 2\sqrt[6]{\frac{b}{c}}+\sqrt[6]{\frac{c}{a}}+\sqrt[6]{\frac{a}{b}}+1\ge 5\cdot\sqrt[30]{\frac{b}{c}}\\ 2\sqrt[6]{\frac{c}{a}}+\sqrt[6]{\frac{a}{b}}+\sqrt[6]{\frac{b}{c}}+1\ge 5\cdot\sqrt[30]{\frac{c}{a}} \end{align}

$\displaystyle 4\sum_{cycl}\sqrt[6]{\frac{a}{b}}+3\ge 5\sum_{cycl}\sqrt[30]{\frac{a}{b}}.$

Thus, suffice it to prove that

$\displaystyle \frac{1}{4}\left(5\sum_{cycl}\sqrt[30]{\frac{a}{b}}-3\right)\ge\sum_{cycl}\sqrt[30]{\frac{a}{b}},$

i.e., $\displaystyle \sum_{cycl}\sqrt[30]{\frac{a}{b}}\ge 3,\,$ which is obvious.

### Solution 3

Take $(x,y,z)=(\sqrt[30]{a},\sqrt[30]{b},\sqrt[30]{c}).\,$ $x,y,z\gt 0.\,$ We have to prove

$\displaystyle (xyz)^5(x^5y^{10}+y^5z^{10}+z^5x^{10})\ge x^9y^{10}z^{11}+y^9z^{10}x^{11}+z^9x^{10}y^{11},$

or, equivalently,

$\displaystyle x^5y^{10}+y^5z^{10}+z^5x^{10}\ge (xyz)(xy^2+yz^2+zx^2).$

Assume $xyz=1.\,$ We have to prove that $\displaystyle \sum_{cycl}(xy^2)^5\ge\sum_{cycl}xy^2,\,$ which is $\displaystyle \sum_{cycl}X^5\ge\sum_{cycl}X,\,$ with $(X,Y,Z)=(xy^2,yz^2,zx^2),\,$ $X,Y,Z\gt 0\,$ and $XYZ=1.$

We have

\displaystyle \begin{align} (X^5+Y^5+Z^5)(X+Y+Z)&\ge(X^3+Y^3+Z^3)^2\;\Rightarrow\\ X^5+Y^5+Z^5&\ge\frac{(X^3+Y^3+Z^3)^2}{X+Y+Z}\\ (X^3+Y^3+Z^3)(X+Y+Z)&\ge(X^2+Y^2+Z^2)^2\\ &\ge\frac{(X+Y+Z)^4}{9}\;\Rightarrow\\ X^3+Y^3+Z^3&\ge\frac{(X+Y+Z)^3}{9}\;\Rightarrow\\ X^5+Y^5+Z^5&\ge\frac{(X^3+Y^3+Z^3)^2}{X+Y+Z}\\ &\ge\frac{(X+Y+Z)^5}{81}\ge X+Y+Z. \end{align}

### Solution 4

Set $x=a^5b^{10}c^{15},\,$ $y=b^5c^{10}a^{15},\,$ $z=c^5a^{10}b^{15}.\,$ Then

\displaystyle \begin{align} 2\sqrt[30]{x}+2\sqrt[30]{y}+\sqrt[30]{z}&\ge 5\sqrt[5]{\sqrt[30]{xxyyz}}=5\sqrt[30]{a^{10}b^{9}c^{11}}\\ 2\sqrt[30]{y}+2\sqrt[30]{z}+\sqrt[30]{x}&\ge 5\sqrt[30]{b^{10}c^{9}a^{11}}\\ 2\sqrt[30]{z}+2\sqrt[30]{x}+\sqrt[30]{y}&\ge 5\sqrt[30]{c^{10}a^{9}b^{11}} \end{align}

It follows that

$\displaystyle \sum_{cycl}\sqrt[6]{ab^2c^3}\ge\sum_{cycl}\sqrt[30]{a^{9}b^{10}c^{11}}.$

### Solution 5

We can rewrite the inequality as $\displaystyle \sum_{cycl}\sqrt[30]{a^5b^{10}c^{15}}\ge\sum_{cycl}\sqrt[30]{a^9b^{10}c^{11}}\,$ and then use convexity arguments.

In general, for $q\gt p_1\gt p_2\gt 0\,$ and $n\gt 0,$

$\displaystyle \sum_{cycl}\sqrt[n]{a^{q-p_1}b^qc^{q+p_1}}\ge \sum_{cycl}\sqrt[n]{a^{q-p_2}b^qc^{q+p_2}}\ge 3a^{\frac{q}{n}}b^{\frac{q}{n}}c^{\frac{q}{n}}.$

### Acknowledgment

Dan Sitaru has kindly posted a problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page by. Solution 1 is by Kevin Soto Palacios (Peru) and, independently, by Ravi Prakash (India); Solution 2 is by Mohammed Jamal (Morocco); Solution 3 is by Nguyen Ngoc Tu (Vietnam); Solution 4 is by Sanong Hauerai (Thailand), Solution 5 is by N. N. Taleb (USA/Lebanon).