# An Inequality in Two Or More Variables II

### Solution 1

Consider function $f:\,[0,\infty)\to\mathbb{R},\,$ defined by $\displaystyle f(x)=(x+1)\ln(x+1)-x-\frac{x^2}{2}.$

\displaystyle \begin{align} &f'(x)=\ln(x+1)+1-1-x=\ln(x+1)-x\\ &f''(x)=\frac{1}{1+x}-1=-\frac{x}{1+x}\lt 0\\ &f''(x)\le f''(0)\,)\Rightarrow f'(x)\leq f'(0)=0\,\Rightarrow f(x)\leq 0, (\forall) x\geq 0\\ &(x+1)\ln (x+1)-x-\frac{x^2}{2}\leq 0\\ &\ln (x+1)^{x+1}\leq x+\frac{x^2}{2}\\ &(x+1)^{x+1}\leq e^{x+\frac{x^2}{2}}; (\forall) x\geq 0 \end{align}

(1)

$(a+1)^{a+1}\leq e^a\cdot \sqrt{e^{a^2}}$

(1)

$(b+1)^{b+1}\leq e^b\cdot \sqrt{e^{b^2}}$

(1)

$(c+1)^{c+1}\leq e^c\cdot \sqrt{e^{c^2}}$

Multiply (1)-(3) to get

$\displaystyle (a+1)^{a+1}\cdot (b+1)^{b+1}\cdot (c+1)^{c+1}\leq e^{a+b+c}\cdot \sqrt{e^{a^2+b^2+c^2}}.$

### Solution 2

Let $x=a+1$, etc. The inequality becomes:

$x^x y^y z^z\leq e^{x+y+z-3} \sqrt{e^{(x-1)^2+(y-1)^2+(z-1)^2}}, \;\; x,y,z>1$

Taking logs on both sides:

$x \log (x)+y \log (y)+z \log (z)\leq\frac{1}{2} \left(x^2+y^2+z^2-3\right)$

Rewriting. We need to minimize

$f(x,y,z)= x^2-2 x \log (x)+y^2-2 y \log (y)+z^2-2 z \log (z)-3$

which is additively separable into $f(x,y,z)=f_1(x)+f_2(y)+f_3(z)$, with $f_1(x)=x^2-2 x \log (x)-1$, etc. The minimum for $f_1(x)$ is for $x=1$, and so on, hence $f(x,y,z)=0$ for $x=y=z=1$, which corresponds to $a=b=c=0$

### Acknowledgment

Dan Sitaru has kindly posted the problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page. He later mailed his solution on a LaTex file - something I appreciate greatly. Solution 2 is by N. N. Taleb.

As the solutions show, the inequality can be extended to any number of variable.