Confusion in proof of inequality I found this question in a Facebook group

Show that, for positive reals $a,b$ and $c$, the following inequality holds:
$\varphi(a,b,c)=\displaystyle \sum_{cyc}\frac{b-a}{\sqrt{(a+\frac1a)}}≥0$

So basically what the author did, is calculate the first and second derivatives, he then said: $\vec \nabla \varphi(1,1,1)=\vec 0$. Thus $(1,1,1)$ is an extremum. Let's determine its nature, we study the Hessian.
The Hessian is given by:
$\textbf{Hess}\varphi(1,1,1)=\begin{pmatrix}
\frac1{\sqrt2} & 0 & 0\\
0 & \frac1{\sqrt2} & 0\\ 0 & 0 & \frac1{\sqrt2}
\end{pmatrix}$
it's positive definite, thus $\varphi$ attains a minimum in $(1,1,1)$.
Finally : $\varphi(a,b,c)≥\varphi(1,1,1)=0$.
My question is:

Doesn't the Hessian tell us only if it's a local minimum or maximum? If yes, how can we show it is a global minimum? Are there any other ways to prove this inequality?

 A: It looks like you have the incorrect Hessian at (1,1,1). I (using the online derivative calculator) calculate it to be given by
\begin{equation}
    \mathbf{Hess} \varphi(1,1,1) = \begin{bmatrix}
        0 & 0 & 0 \\
        0 & 0 & 0 \\
        0 & 0 & 0
    \end{bmatrix}
\end{equation}
which is not useful for the second derivative test.
Stepping back a bit though, you are right. In general, the Hessian $\mathbf{Hess} \varphi$ at a single point $(a,b,c)$ can only tell you if the point is a local minimum or maximum (this is the second derivative test).
However, if the Hessian $\mathbf{Hess} \varphi$ is positive semi-definite everywhere on a convex set $\Omega$ (for example $\mathbb{R}^3_+$ in your case), then $\varphi$ is a convex function on $\Omega$ (this is the second order characterization of convexity). For a convex function every local minimum is a global minimum.
Their proof has fallen short of showing $\mathbf{Hess} \varphi$ is positive semi-definite everywhere and as it stands is not sufficient to prove the result.
Just to wrap up the specific example, if you choose $b = c = 1$ then
\begin{align}
   \varphi(a,1,1) &= \frac{1-a}{\sqrt{a+1/a}} + \frac{1-1}{\sqrt{1+1/1}} + \frac{a-1}{\sqrt{1+1/1}} \\
&= \frac{1-a}{\sqrt{a+1/a}} + \frac{a-1}{\sqrt{2}}
\end{align}
and choosing $a = 1/2$ gives
\begin{equation}
   \varphi(1/2,1,1) = \frac{1-1/2}{\sqrt{1/2 + 2}} + \frac{1/2-1}{\sqrt{2}} = \frac{1}{2\sqrt{5/2}} - \frac{1}{2\sqrt{2}} < 0 
\end{equation}
so the claim doesn't seem to be true either
