# Hlawka-like Dinca's Inequality

The subject of the page is to establish an analogue of Hlawka's inequality by Daniel Sitaru, Dan Marinescu and Leo Giugiuc. Marian Dinca appears to have proved a more general statement.

Note that the original statement by Daniel Sitaru, Dan Marinescu and Leo Giugiuc dealt only with the case $\alpha\in (0,1).\;$ At the artofproblemsolving forum that inequality has been stated for $\alpha\;$ an integer.

### Proof for $\alpha\in (0,1)$

Define $f(x)=(x)^{\alpha}+(y)^{\alpha}+(z)^{\alpha}+(x+y+z)^{\alpha}-(x+y)^{\alpha}-(y+z)^{\alpha}-(z+x)^{\alpha}.\;$ Then $f'(x)=\alpha x^{\alpha -1}+\alpha (x+y+z)^{\alpha -1}-\alpha (x+y)^{\alpha -1}-\alpha (z+x)^{\alpha -1}.$ Now observe that $x\le x+y\le x+y+z\;$ which implies the existence of $\lambda\in [0,1]\;$ such that $x+y=\lambda x+(1-\lambda )(x+y+z).\;$ Similarly, there is $\mu\in [0,1]\;$ such that $x+z=\mu x+(1-\mu )(x+y+z).\;$ It follows that $(x+y)+(x+z)=(\lambda+\mu )+(2-\lambda-\mu)(x+y+z),\;$ or

$y+z=(\lambda+\mu-2)x+(2-\lambda-\mu)(x+y+z)=(2-\lambda-\mu)(y+z).$

In other words, $(1-\lambda-\mu)(y+z)=0.\;$ If $y+z=0\;$ then $y=z=0\;$ and the required inequality becomes $x^{\alpha}+x^{\alpha}\le x^{\alpha}+x^{\alpha},\;$ which is certainly true. Thus assume $\lambda+\mu=1.\;$

For $\alpha\in (0,1),\;$ unction $g(t)=t^{\alpha-1}\;$ is convex on $(0,1).\;$ Using Jensen's inequality,

$g(x+y)=g(\lambda x+(1-\lambda )(x+y+z)\le \lambda g(x)+(1-\lambda)g(x+y+z),\\ g(x+z)=g(\mu x+(1-\mu )(x+y+z)\le \mu g(x)+(1-\mu)g(x+y+z)$

such that

\begin{align} g(x+y)+g(x+z)&\le (\lambda+\mu)g(x)+(2-\lambda-\mu)g(x+y+z)\\ &=g(x)+g(x+y+z), \end{align}

implying $f'(x)\ge 0\;$ and, subsequently, $f(x)\ge f(0)=0.$

### Proof for $\alpha\ge 1$

Function $f(t)=t^{\alpha}\;$ is convex, lending itself to an application of the Hardy-Littlewood-Polya inequality for the sequences $\{x+y+z,x,y,z\}\;$ and $\{x+y,y+z,z+x,0\}.\;$ Assuming $x\ge y\ge z,\;$ we obtain

$f(x+y+z)+f(x)+f(y)+f(z)\ge f(x+y)+f(y+z)+f(z+x).$

### Proof for $\alpha\le 0$

In this case $x^{\alpha}\ge (x+y)^{\alpha},\;$ $y^{\alpha}\ge (y+z)^{\alpha},\;$ $z^{\alpha}\ge (z+x)^{\alpha}.\;$ Summing up,

\begin{align} \;x^{\alpha}+y^{\alpha}+z^{\alpha}+(x+y+z)^{\alpha}&\ge x^{\alpha}+y^{\alpha}+z^{\alpha}\\ &\ge (x+y)^{\alpha}+(y+z)^{\alpha}+(z+x)^{\alpha}. \end{align}

### Counterexample

Leo Giugiuc informed me of the following counterexample: let $\displaystyle\alpha=\frac{3}{2},\;$ $x=y=z=1.\;$ Then

\begin{align} \;x^{\alpha}+y^{\alpha}+z^{\alpha}+(x+y+z)^{\alpha}&=3+3\sqrt{3}\\ &\approx 8.19615\\ &\lt 8.48528\\ &\approx 6\sqrt{2}\\ &=3\cdot 2^3/2\\ &=(x+y)^{\alpha}+(y+z)^{\alpha}+(z+x)^{\alpha}. \end{align}