# A Cyclic Inequality in Three Variables And One More

### Solution 1

Lemma 1 (Iran 1996)

If $a,b,c\ge 0\,$ then $\displaystyle(ab+bc+ca)\left(\sum_{cycl}\frac{1}{(a+b)^2}\right)\ge\frac{9}{4}.$

Lemma 2 Function $f(t)=t^{m+1}\,$ is convex on $(0,\infty).$

Both results are well know and left as an exercise to the reader.

From Lemma 2, $(y+z)^{m+1}\le 2^m(y^{m+1}+z^{m+1})\,$ implying

$\displaystyle\frac{1}{(y+z)^{2m+2}}\ge\frac{1}{4^m(y^{m+1}+z^{m+1})^2}.$

Let $x^{m+1}=a,\,$ $y^{m+1}=b\,$ and $z^{m+1}=c.\,$ Then $\displaystyle\sum_{cycl}x^{2m+2}=\sum_{cycl}a^2\ge ab+bc+ca\;$ and, from the above,

$\displaystyle\sum_{cycl}\frac{1}{(y+z)^{2m+2}}\ge\frac{1}{4^m}\sum_{cycl}\frac{1}{(y^{m+1}+z^{m+1})^2}=\frac{1}{4^m}\left(\sum_{cycl}\frac{1}{(a+b)^2}\right).$

But from Lemma 1, $\displaystyle(ab+bc+ca)\left(\sum_{cycl}\frac{1}{(a+b)^2}\right)\ge\frac{9}{4}.$

### Solution 2

By the Cauchy-Schwarz inequality,

$\displaystyle\sum_{cycl}x^{2m+2}\cdot\sum_{cycl}\frac{1}{(y+z)^{2m+2}}\ge\left(\sum_{cycl}\frac{x^{m+1}}{(y+z)^{m+1}}\right)^2.$

Suffice it to show that $\displaystyle\sum_{cycl}\frac{x^{m+1}}{(y+z)^{m+1}}\ge\frac{3}{2^{m+1}}.$

Now, by the power mean,

$\displaystyle\sum_{cycl}\frac{1}{(y+z)^{m+1}}\ge\frac{1}{2^{m}(y^{m+1}+z^{m+1})}.$

Thus, we need only to show that $\displaystyle\sum_{cycl}\frac{x^{m+1}}{y^{m+1}+z^{m+1}}\ge\frac{3}{2}.$

Now assume, WLOG, $x\ge y\ge z.\,$ Then

$\displaystyle\frac{1}{y^{m+1}+z^{m+1}}\ge\frac{1}{z^{m+1}+x^{m+1}}\ge\frac{1}{x^{m+1}+y^{m+1}}.$

By the Rearrangement inequality, $\displaystyle\sum_{cycl}\frac{x^{m+1}}{y^{m+1}+z^{m+1}}\ge\sum_{cycl}\frac{x^{m+1}}{z^{m+1}+x^{m+1}}\,$ as well as $\displaystyle\sum_{cycl}\frac{x^{m+1}}{y^{m+1}+z^{m+1}}\ge\sum_{cycl}\frac{z^{m+1}}{z^{m+1}+x^{m+1}}.\,$ Adding up we get $\displaystyle\sum_{cycl}\frac{x^{m+1}}{y^{m+1}+z^{m+1}}\ge\frac{3}{2},\,$ as expected.

### Solution 3

\displaystyle\begin{align} LHS &\overset{CBS}{\ge}\left(\sum_{cycl}\left(\frac{x}{y+z}\right)^{m+1}\right)^2\;\overset{generalized\\means}{\ge}\;\left(3\left(\frac{\displaystyle\sum_{cycl}\frac{x}{y+z}}{3}\right)^{m+1}\right)^2\\ &\overset{Nesbitt's}{\ge}\left(3\left(\frac{3/2}{3}\right)^{m+1}\right)^2=\frac{9}{4^{m+1}}. \end{align}

### Solution 4

\displaystyle\begin{align} &S_1=\sum_{cycl}x^{2m+2}=\sum_{cycl}a^2,\\ &\left(\frac{y+z}{2}\right)^{m+1}\le\frac{y^{m+1}+z^{m+1}}{2}=\frac{b+c}{2},\text{so that},\\ &S_2=\left(\sum_{cycl}\frac{1}{(y+z)^{2m+2}}\right)\ge\sum_{cycl}\left(\frac{1}{2^m}\cdot\frac{1}{b+c}\right)^2\\ &\qquad\qquad\overset{Bergstrom's}{\ge}\frac{1}{4^m}\left(\frac{3^2}{\displaystyle\sum_{cycl}(a+b)^2}\right)\ge\frac{1}{4^m}\cdot\frac{9}{\displaystyle 4\sum_{cycl}a^2}. \end{align}

Now, $S_1\cdot S_2\ge\displaystyle\frac{9}{4^{m+1}}.$

### Acknowledgment

The problem from the Romanian Mathematical Magazine has been kindly posted at the CutTheKnotMath facebook page by Leo Giugiuc, along with his solution (Solution 1). (The problem is due to D. M. Batinetu-Giurgiu and Neculai Stanciu.) Dan Sitaru Has later added proofs by Saptak Bhattacharya (Solution 2) and two solutions by Daniel Dan (Solutions 3 and 4.)