# A Partly Cyclic Inequality in Four Variables

### Solution

Let $f(x)=xe^x, f''(x)=(x+2)e^x.\,$ Thus for $x\ge -2,\,$ $f\,$ is convex so that, by Jensen's inequality,

$\displaystyle \frac{f(z)+f(t)+f(-2)}{3}\ge f\left(\frac{z+t-2}{3}\right),$

i.e.,

(1)

$\displaystyle f(z)+f(t)+f(-2)\ge 3f\left(\frac{z+t-2}{3}\right).$

For $x\le -2,\,$ $f(x)\,$ is concave. $x+y=-2+(x+y+2).\,$ Thus, by Karamata's inequality,

(2)

$\displaystyle f(x)+f(y)+f(-2)\ge f(-2)+f(x+y+2).$

Adding up (1) and (2) gives

$\displaystyle f(z)+f(t)+f(-2)+f(x)+f(y)\ge f(-2)+f(x+y+2)+3f\left(\frac{z+t-2}{3}\right),$

i.e.,$\displaystyle \sum_{cycl}f(t)\ge f(x+y+2)+3f\left(\frac{z+t-2}{3}\right)\,$ or more explicitly,

$\displaystyle ze^z+te^t+xe^x+ye^y\ge (x+y+2)e^{x+y+2}+3\frac{z+t-2}{3}\sqrt[3]{e^{z+t-2}},$

and, finally,

$\displaystyle ze^z+te^t+xe^x+ye^y\ge (x+y+2)e^{x+y+2}+(z+t-2)\sqrt[3]{e^{z+t-2}}.$

### Acknowledgment

Dan Sitaru has kindly posted the problem from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later communicated the above solution in a LaTeX file.