$n>0$, $x,y,z>0$ are arbitrary real numbers.
By Hölder's inequality
$$\tag1
(a+b+c)^{1\over n+1}(a x^{n + 1} + b y^{n + 1} + c z^{n + 1})^{n\over n+1}\ge ax^n+by^n+cz^n$$
so
Given $ax^n+by^n+cz^n=a+b+c$, the mininum of $a x^{n + 1} + b y^{n + 1} + c z^{n + 1}$ is $a+b+c$.
When $n\to0$, using $\lim\frac{x^n-1}{n\ln(x)}=1$, the condition$$ax^n+by^n+cz^n=a+b+c⇔a(x^n-1)+b(y^n-1)+c(z^n-1)=0$$is close to $a\ln(x)+b\ln(y)+c\ln(z)=0$.
From the above
Given $a\ln(x)+b\ln(y)+c\ln(z)=0$, the mininum of $a x+ b y+ c z$ is $a+b+c$.
Is this true?
My attempt:
Dividing (1) by $a+b+c$, raising to a power of $\frac1n$
$$\tag2\left(a x^{n + 1} + b y^{n + 1} + c z^{n + 1}\over a+b+c\right)^{1\over n+1}\ge\left(\frac{ax^n+by^n+cz^n}{a+b+c}\right)^{1\over n}$$
The right-hand side equals
$$\left(1+\frac{a(x^n-1)+b(y^n-1)+c(z^n-1)}{a+b+c}\right)^{1\over n}$$
When $n\to0$, this is close to
$$\left(1+n\frac{a \ln (x)+b \ln (y)+c \ln (z)}{a+b+c}\right)^{\frac1n}$$
Using $\lim_{n\to0}(1+nx)^{\frac1n}=e^x$, this equals
$$\exp\left(\frac{a \ln (x)+b \ln (y)+c \ln (z)}{a+b+c}\right)$$
so $(2)$ becomes
$${a x + b y + c z\over a+b+c}\ge\exp\left(\frac{a \ln (x)+b \ln (y)+c \ln (z)}{a+b+c}\right)$$
This proves the claim.
Original question:
given $a\ln(x)+b\ln(y)+c\ln(z)=0$, the mininum of $a x^2+ b y^2+ c z^2$ is $a+b+c$.
This can be proved using same method:
By Hölder's inequality
$$\left(a x^{n + 2} + b y^{n + 2} + c z^{n + 2}\right)^n(a+b+c)^2\ge\left(ax^n+by^n+cz^n\right)^{n+2}$$
so
$$\left(a x^{n + 2} + b y^{n + 2} + c z^{n + 2}\over a+b+c\right)^{1\over n+2}\ge\left(\frac{ax^n+by^n+cz^n}{a+b+c}\right)^{1\over n}$$
When $n→0$, using $\lim\frac{x^n-1}{n\ln(x)}=1$,
$$\left(a x^2 + b y^2 + c z^2\over a+b+c\right)^{1\over2}\ge\exp\left(\frac{a \ln (x)+b \ln (y)+c \ln (z)}{a+b+c}\right)$$
This proves the claim.