<s>[INST] Minimize $f(a,b,c) = \dfrac{1}{a²+b²+c²} +\dfrac{1}{9abc}$ subject to $a+b+c=1$ and $a,b,c>0$ 
Minimize $f(a,b,c) = \dfrac{1}{a²+b²+c²} +\dfrac{1}{9abc}$ subject to
$a+b+c=1$ and $a,b,c>0$

My attempt:
Since $f(a,b,c) \ge 0$, then we can minimizing the addends individually. Furthermore, minimizing fractions is the same as maximizing their denominators.
Apply the Lagrange Multiplier on ${a²+b²+c²}$ with the contraints $\implies a=b=c=\frac{1}{3}.$
Similarly, apply the Lagrange Multiplier on ${9abc}$ with the contraints $\implies a=b=c=\frac{1}{3}.$
Therefore, the minimum of $f\!\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)=6.$
I'm looking for a better solution not using the Lagrange Multiplier.
 [/INST] Let's try this with AM-GMs. First note that as $(x+y+z)^2\geqslant 3(xy+yz+zx)$,
$$(ab+bc+ca)^2 \geqslant 3abc(a+b+c)=3abc $$
$\begin{align*}
 \implies 9abc(a^2+b^2+c^2) & \leqslant 3(a^2+b^2+c^2)(ab+bc+ca)^2 \\
 &\leqslant 3\left(\frac{a^2+b^2+c^2+2(ab+bc+ca)}3 \right)^3 \\
&=\frac19(a+b+c)^6=\frac19
\end{align*}$
Hence
$$f \geqslant \frac2{\sqrt{(a^2+b^2+c^2)(9abc)}}\geqslant \frac2{1/3}=6$$
with equality iff $a=b=c=\frac13$.
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