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

Question

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.

Answer 1

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 $a=b=c=\frac13$.

Answer 2

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 $$(1/(a^2+b^2+c^2))+(1/(9abc))-6≥(1/(1-2t))+(1/(3t^2))-6=$$ $$((4t+1)(1-3t)^2)/(3t^2(1-2t))≥0.$$