<s>[INST] Prove $\dfrac{1}{x} +\dfrac{1}{y} +\dfrac{1}{z} -\dfrac{1}{x+y} - \dfrac{1}{y+z} -\dfrac{1}{x+z} +\dfrac{1}{x+y+z}$ is a convex function My problem: Prove $g(x,y,z)=\dfrac{1}{x} +\dfrac{1}{y} +\dfrac{1}{z} -\dfrac{1}{x+y} - \dfrac{1}{y+z} -\dfrac{1}{x+z} +\dfrac{1}{x+y+z}$ with $x,y,z >0$ is a convex function.
I tried to use the definition of convex function. Moreover, i tried to use the differentiating but i still get stuck with them.
Can you help me to solve this problem.
 [/INST] Hints.
If $f(x, y, z): \Omega \to \mathbb{R}$ is twice differentiable and the Hessian is positive semidefinite in the entire domain $\Omega$, then $f(x, y, z)$ is convex.
$\Omega$ must be assumed to be convex too.
If the Hessian has a negative eigenvalue at a point $(x_0, y_0, z_0) \in\Omega$, then the function is not convex.
The domain in your case is convex, since $x>0$, $y>0$, $z>0$
Can you proceed?
THe Hessian Matrix reads:
$$\left(
\begin{array}{ccc}
 \frac{2}{x^3}+\frac{2}{(x+y+z)^3}-\frac{2}{(x+y)^3}-\frac{2}{(x+z)^3} & \frac{2}{(x+y+z)^3}-\frac{2}{(x+y)^3} & \frac{2}{(x+y+z)^3}-\frac{2}{(x+z)^3} \\
 \frac{2}{(x+y+z)^3}-\frac{2}{(x+y)^3} & \frac{2}{(x+y+z)^3}-\frac{2}{(x+y)^3}+\frac{2}{y^3}-\frac{2}{(y-z)^3} & \frac{2}{(x+y+z)^3}+\frac{2}{(y-z)^3} \\
 \frac{2}{(x+y+z)^3}-\frac{2}{(x+z)^3} & \frac{2}{(x+y+z)^3}+\frac{2}{(y-z)^3} & \frac{2}{(x+y+z)^3}-\frac{2}{(x+z)^3}-\frac{2}{(y-z)^3}+\frac{2}{z^3} \\
\end{array}
\right)$$
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