Prove that an element $(a,b,c)$ belongs to this set such that it satisfies a certain condition This particular question was asked in a masters entrance for which I am preparing.

Let $E =\{(x,y,z)\in \mathbb{R}^{3}\ |\ x,y,z >0 , xy+yz+zx=1 \}$. Prove that there exists $(a,b,c)\in E$ such that $\space abc\geq xyz \space$ for all $ \space(x,y,z)\in E$.

I have no idea what branch of mathematics should be used to solve it and so I can't provide anything in attempt.
Kindly guide me.
 A: We just use AM-GM inequality.
$$(xyz)^{\frac{2}{3}}\leq\frac{xy+yz+zx}{3}=\frac{1}{3}\implies xyz\leq \frac{1}{3\sqrt{3}}$$ with equality if and only if $xy=yz=zx=\frac{1}{3}\implies x=y=z=\frac{1}{\sqrt{3}}$. So we can take $(a,b,c)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$. Done!
A: There may be an easier way to prove this, but it looks like a constrained optimization problem to me. You could reframe the problem in the following way:
Prove that $f(x,y,z) = xyz $ has a maximum when $(x,y,z)$ are subject to the constraint $h(x,y,z) = 0$, where $h(x,y,z) = xy + yz + zx - 1$.
Then, to prove it you could appeal to the KKT theorem (see https://en.wikipedia.org/wiki/Karush–Kuhn–Tucker_conditions), which in this case is just the idea of optimization using Lagrange multipliers.
First you might use the first-order necessary condition from KKT to find $(a,b,c)$:
$$ Df(a,b,c) - \lambda^* Dh(a,b,c) = 0, ~\text{subject to}~ h(a,b,c) = 0 .$$
Then, you can complete the proof by showing that the second-order sufficient condition is satisfied by $(a,b,c)$. Letting $$\mathscr{L}(x,y,z,\lambda) := f(x,y,z) - \lambda h(x,y,z), $$ you need to show $$ v^T D^2_\mathbf{x} \mathscr{L} (a,b,c,\lambda^*) v > 0 $$
for all $ v $ in the null space of $Dh(a,b,c)$.
Note: in my confusing notation for $D^2_\mathbf{x}  \mathscr{L}$, I mean that we only take the total derivative with respect to the arguments $(x,y,z)$, not with respect to $\lambda$.
