Mathematical Reflections 2012, Issue 6 - Problem S251

Problem:
Find all triples $(x,y,z)$ of positive real numbers for which there is a positive real number $t$ such that the following inequalities hold simultaneously:
$$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+t\leq4, \qquad x^2+y^2+z^2+\dfrac{2}{t}\leq5.$$

Proposed by Titu Andreescu.

Solution:
From the two inequalities, it's easy to see that if there exists such a positive real number $t$, then $t$ satisfies
\begin{equation}\label{first}
\dfrac{2}{5}<t<4.                                                              (1)
\end{equation}
From the first inequality, using the AM-HM inequality we obtain $$\dfrac{9}{4-t} \leq \dfrac{9}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \leq x+y+z.$$
Now, suppose without loss of generality that $x \geq y \geq z$. Then, $x^2 \geq y^2 \geq z^2$ and $1/x \leq 1/y \leq 1/z$.
By Chebyshev's Inequality, we have $$x+y+z \leq \dfrac{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x^2+y^2+z^2)}{3} \leq \dfrac{(5-2/t)(4-t)}{3}.$$ Hence, $$\dfrac{9}{4-t} \leq \dfrac{(5-2/t)(4-t)}{3}$$ and after simple calculations we get
$$5t^3-42t^2+69t-32 \geq 0 \iff (t-1)^2(5t-32) \geq 0,$$ which gives
\begin{equation}\label{second}
t=1, \qquad t \geq \dfrac{32}{5}.                                       (2)
\end{equation}
From $(1)$ and $(2)$, we get $t=1$, from which $x+y+z=1/x+1/y+1/z=3$. But this implies that the arithmetic mean and harmonic mean are equal, so $x=y=z=1$ and the only triple is $(1,1,1)$.