15th Czech-Polish-Slovak Mathematics Competition

Solution: By changing $(x,y,z)$ to $(-x,-y,-z)$, the condition (1) remains valid and the value of $x+y+z$ changes sign. Since the ordering of the numbers $x,y,z$ is irrelevant and they can not be of the same sign because of (1), we can without the loss of generality assume $x>0,y>0$, and $z<0$, and find the maximum value of $V=|x+y+z|$. We can transform (1) equivalently to $$f(x)+f(y)=f(t),\text{where}t:=-z>0\text{and}f(x)=x+\frac{1}{x},$$ with $x,y,t\in I:=(1,\infty )$. In the following, we shall use the well known fact about the function $f$: It is an increasing bijection mapping $I$ onto $(2,\infty )$ (the trivial proof is omitted here). Applying this, from $$f(t)=f(x)+f(y)=x+\frac{1}{x}+y+\frac{1}{y}>x+y+\frac{1}{x+y}=f(x+y)$$ we get $t>x+y$, hence $x+y+z=x+y-t<0$. Therefore our aim is to maximize the positive expression $V=|x+y+z|=t-x-y$. We will call valid every triple $(x,y,t)\in I^3$ satisfying $f(x)+f(y)=f(t)$. One of the valid triples is $(1,1,t_0)$, with $t_0>1$ such that $f(t_0)=4$, i. e., $t_0=2+\sqrt{3}$. For this triple, we have $V=\sqrt{3}$. We shall show this is the desired maximum. The proof will be based on the following lemma: Whenever $(x,y,t)$ and $(x,y^{\prime },t^{\prime })$ are valid triples with the same first component $x$, and $y^{\prime }<y$, we have $t-x-y<t^{\prime }-x-y^{\prime }$. If this lemma is true, then, with respect to the symmetry, the similar conclusion holds for the valid triples $(x,y,t)$ and $(x^{\prime },y,t^{\prime })$ with the same second component $y$. Hence, any valid triple $(x,y,t)$ can be replaced by $(x,1,t^{\prime })$ and then by $(1,1,t_0)$, and the value of $V$ increases or remains the same during this process, concluding $\max V=\sqrt{3}$. (Moreover, this also implies that $(1,1,2+\sqrt{3})$ is the only valid triple for which the maximum value $V=\sqrt{3}$ is reached.) To prove the lemma, notice that from $$f(t)=f(x)+f(y),f(t^{\prime })=f(x)+f(y^{\prime }),$$ and from $f(y^{\prime })<f(y)$ (which is implied by the given assumption $y^{\prime }<y$), we have $f(x)<f(t^{\prime })<f(t)$, hence $1<t^{\prime }<t$. Therefore $1<t^{\prime }{y}^{\prime }<ty$, and from $$f(x)=f(t)-f(y)=(t-y)\left(1-\frac{1}{ty}\right)=f(t^{\prime })-f(y^{\prime })=(t^{\prime }-y^{\prime })\left(1-\frac{1}{t^{\prime }{y}^{\prime }}\right),$$ using the estimates $$0<1-\frac{1}{t^{\prime }{y}^{\prime }}<1-\frac{1}{ty},$$ we obtain $t-y<t^{\prime }-y^{\prime }$, which is equivalent to the desired $t-x-y<t^{\prime }-x-y^{\prime }$. Answer. The maximum value of $x+y+z$ is $\sqrt{3}$.

---

Alternative solution.

Second solution. We start in the same way as in the previous solution, reducing to the case $x\ge 1$, $y\ge 1$, and $z\le -1$ with maximizing $V=|x+y+z|$. From (1) we can express $z$ as the solution of the quadratic equation $$z^2+\left(x+\frac{1}{x}+y+\frac{1}{y}\right)z+1=0.$$ Since the product of the roots of this equation equals 1, the value of $z$ lying outside of $(-1,1)$ is, with respect to $x\ge 1$, $y\ge 1$, given by $$z=\frac{1}{2}\left(-\left(x+\frac{1}{x}+y+\frac{1}{y}\right)-\sqrt{{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)}^2-4}\right).$$ Therefore we are left to maximize $$2V=2|-x-y-z|=\left|\sqrt{{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)}^2-4}-\left(x-\frac{1}{x}+y-\frac{1}{y}\right)\right|.$$ One can easily check that the expression inside of the last bars is positive. It remains to show that $$\sqrt{{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)}^2-4}-\left(x-\frac{1}{x}+y-\frac{1}{y}\right)\le 2\sqrt{3}.$$ This inequality is equivalent (after rearranging, checking the positivity of both sides, canceling the square root, and simple calculations) to $$\begin{gathered} \sqrt{{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)}^2-4}\le 2\sqrt{3}+\left(x-\frac{1}{x}+y-\frac{1}{y}\right), \\ {x}^2+y^2+\sqrt{3}x+\sqrt{3}y\le 2xy+\sqrt{3}{x}^{2}y+\sqrt{3}x{y}^2, \end{gathered}$$ which is true, since it is the sum of the inequalities $$x^2\le x^{2}y,y^2\le x{y}^2,\sqrt{3}x\le (\sqrt{3}-1)x^{2}y+xy,\text{and}\sqrt{3}y\le (\sqrt{3}-1)x{y}^2+xy,$$ and these are trivially valid for $x,y\ge 1$.