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Let $x$, $y$, $z$ be positive real numbers and $x + y + z =3$. Does a maximum value exist for this expression? $$\displaystyle E = \frac{x}{2 y+3 z}+\frac{4 y}{5 z + 6 x}+\frac{7 z}{8x+9 y}.$$ I tried Put $$a=2 y+3 z,\quad b=6 x+5 z,\quad c=8 x+9y.$$ Then $$x = -\dfrac{45a}{242}+\dfrac{27b}{242}+\dfrac{5c}{121},\quad y = \dfrac{20a}{121}-\dfrac{12b}{121}+\dfrac{9c}{121}, \quad z = \dfrac{27a}{121}+\dfrac{8b}{121}-\dfrac{6c}{121}.$$ The expression $E$ has the form $$E = \frac{378 a^2 b+160 a^2 c+112a b^2-225 a b c+72 a c^2+27b^2 c+10 b c^2}{242 a b c}.$$ From here, I can't solve it.

minthao_2011
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1 Answers1

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The expression $E$ is not bounded above. Let $x$ be very large, and $y$ and $z$ small.

Edit: The condition $x+y+z=3$ has been added. The expression $E$ is still not bounded above. Let $x$ be very close to $3$, making $y$ and $z$ close to $0$. The term $\frac{x}{2y+3z}$ can be made arbitrarily large.

André Nicolas
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