<s>[INST] the $\max$ of $\min$ $a,b,c>0$,
$$\begin{cases}a+b+c=8\\ab+bc+ac=16\end{cases}$$
$m=\min\{ab,bc,ca\}$ , how to get $m_{\max}$
With Mathematica
Maximize[{Min[a b, b c, c a], a + b + c == 8, a b + b c + c a == 16, a > 0, b 
> 0, c > 0}, {a, b, c}]

$$\left\{\frac{16}{9},\left\{a\to \frac{4}{3},b\to \frac{4}{3},c\to \frac{16}{3}\right\}\right\}$$
but how to do it by hand?
 [/INST] Let $a\geqslant b\geqslant c$, $b=c+u$ and $a=c+u+v$.
Thus, $u$ and $v$ are non negatives, $$3c+2u+v=8$$ and
$$(c+u)(c+u+v)+c(c+u)+c(c+u+v)=16$$ or
$$3c^2+(4u+2v)c+u(u+v)=16$$ or
$$9c^2+(4u+2v)3c+3u(u+v)=48,$$ which gives
$$(8-2u-v)^2+(4u+2v)(8-2u-v)+3u^2+3uv=48$$ or
$$u^2+v^2+uv=16.$$
Now, since $$ab\geqslant ac\geqslant bc,$$ we'll prove that
$$c(c+u)\leqslant\frac{16}{9}$$ or
$$\frac{8-2u-v}{3}\left(\frac{8-2u-v}{3}+u\right)\leqslant\frac{16}{9}$$ or
$$2u^2-uv-v^2+8(u+2v)\geqslant48$$ or
$$2u^2-uv-v^2+2(u+2v)\sqrt{u^2+uv+v^2}\geqslant3(u^2+uv+v^2)$$ or
$$2(u+2v)\sqrt{u^2+uv+v^2}\geqslant(u+2v)^2$$ or
$$4u^2+4uv+4v^2\geqslant u^2+4uv+4v^2,$$ which is obvious.
The equality occurs for $u=0$ and $v^2=16,$ id est for $(u,v)=(0,4)$, $c=b=\frac{4}{3}$ and $a=\frac{16}{3},$
which says that $$m_{\max}=\frac{16}{9}.$$
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