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Tuesday, March 5, 2013

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).

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