Dan Sitaru's Cyclic Inequality in Four Variables


Dan  Sitaru's  Cyclic  Inequality in Four Variables

Solution 1

First employing Bergstrom's and then the AM-GM inequalities, we get

$\displaystyle \begin{align} LHS&=\sum_{cycl}\frac{a^8}{abcd+a^4} \ge \frac{(a^4+b^4+c^4+d^4)^2}{4abcd+(a^4+b^4+c^4+d^4)}\\ &\ge \frac{(a^4+b^4+c^4+d^4)^2}{(a^4+b^4+c^4+d^4)+(a^4+b^4+c^4+d^4)}\\ &=\frac{(a^4+b^4+c^4+d^4)}{2}\ge\frac{4\sqrt[4]{a^4b^4c^4d^4}}{2}=2abcd=RHS. \end{align}$

Solution 2

The required inequality is homogeneous, letting as assume $abcd=1.$ So suffice it to show that, under that constraint, $\displaystyle \sum_{cycl}\frac{a^7}{bcd+a^3}\ge 2$ which is equivalent to $\displaystyle \sum_{cycl}\frac{a^7}{\displaystyle \frac{1}{a}+a^3}\ge 2.$

Consider $\displaystyle \frac{x^7}{\displaystyle \frac{1}{x}+x^3}=\frac{x^8}{1+x^4},$ $x\gt 0.$

$\displaystyle f'(x)=\frac{4x^7(x^4+2)}{(x^4+1)^2}\gt 0,$ $x\gt 0,$

implying that $f(x)$ is increasing on $(0,\infty).$ Thus, from $\displaystyle \frac{a+b+c+d}{4}\ge\sqrt[4]{abcd},$ we conclude that

$\displaystyle f\left(\frac{a+b+c+d}{4}\right)\ge f(\sqrt[4]{abcd})=f(1)=\frac{1}{2}.$

To continue,

$\displaystyle f''(x)=\frac{4x^6(3x^8+9x^4+14)}{(x^4+1)^3}\gt 0,$

so that $f(x)$ is convex on $(0,\infty).$ By Jensen's inequality then,

$\displaystyle \begin{align} LHS&=\sum_{cycl}\frac{a^7}{\displaystyle \frac{1}{a}+a^3}\ge 4f\left(\frac{a+b+c+d}{2}\right)\\ &=4f(1)=4\cdot\frac{1}{2}=2. \end{align}$

Solution 3

The inequality being homogenoeus, WLOG, let $abcd=1$. Also, let $p=a^4$, $q=b^4$, $r=c^4$, and $s=d^4$. Note, $pqrs=1$. The inequality becomes

$\displaystyle \sum_{cycl} \frac{p^2}{1+p}\geq 2.$

Let, $\displaystyle x=\frac{p+q+r+s}{4}.$ The AM-GM inequality gives $x\geq 1$. For positive arguments, the function $p^2/(1+p)$ is convex. Applying Jensen's inequality,

$\displaystyle \sum_{cycl} \frac{p^2}{1+p} \geq \frac{4x^2}{1+x}.$

For positive arguments, the RHS is a monotonically increasing function of $x$ and takes minimum value when $x$ takes its minimum value of $1$. Thus, $RHS\geq 2$.

Solution 4

By Cauchy-Schwarz inequality we have,

$\displaystyle \begin{align} \left ( \sum_{cyc}^{ }\frac{a^7}{bcd+a^3} \right )\left ( \sum_{cyc}^{ }a(bcd+a^3) \right ) \ge (\sum_{cyc}^{ }a^4)^2\\ \Rightarrow \sum_{cyc}^{ }\frac{a^7}{bcd+a^3} \ge \frac{(\displaystyle \sum_{cyc}^{ }a^4)^2}{\displaystyle \sum_{cyc}^{ }a(bcd+a^3) } \end{align}$

We need to prove}

$\displaystyle \frac{(\sum_{cyc}^{ }a^4)^2}{ \sum_{cyc}^{ }a(bcd+a^3) } \ge 2abcd.$

The last inequality equivalent to

$\displaystyle \left( {a}^{4}+{b}^{4}+{c}^{4}+{d}^{4}-4abcd \right) \left( {a}^{4 }+{b}^{4}+{c}^{4}+{d}^{4}+2\,bcda \right) \ge 0.$


Dan Sitaru has kindly posted problem of his at the CutTheKnotMath facebook page, with several solutions. The problem appeared earlier at the Romanian Mathematical Magazine. Solution 1 is by Rozeta Atanasova; Solution 2 is by Dimitris Kastriotis; Solution 3 is by Amit Itagi; Solution 4 is by Tạ Hỡng Quảng VT. There are more solutions at the page.


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