<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] This solution was posted by Arkady Alt on LinkedIn. Let $t:=ab+bc+ca.$ Since $3abc=3abc(a+b+c)\leq(ab+bc+ca)^2=t^2$,
$a^2+b^2+c^2=1-2t$ and $3t\leq(a+b+c)^2=1$ then
$$(1/(a^2+b^2+c^2))+(1/(9abc))-6≥(1/(1-2t))+(1/(3t^2))-6=$$
$$((4t+1)(1-3t)^2)/(3t^2(1-2t))≥0.$$
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