Question: Prove that the minimum values of $x^2+y^2+z^2$ is $27$, where $x,y,z$ are positive real variables satisfying the condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$.
From AM$\ge$ GM, we have $\left( \dfrac{x^2+y^2+z^2}{3}\right)^3\ge (xyz)^2=(xy+yz+zx)^2$. Is it possible to show thae result from this relation?
By Holder, $$\left(x^2+y^2+z^2 \right)\cdot \left(\frac1x+\frac1y+\frac1z \right)^2\geqslant (1+1+1)^3$$
As $x=y=z=3$ achieves this, we have a minimum.
This is a direct application of harmonic vs. quadratic mean (HM-QM):
Now, plugin $\frac 1x + \frac 1y + \frac 1z = 1$ and you get
$$3 \leq \sqrt{\frac{x^2+y^2+z^2}{3}} \Leftrightarrow 27 \leq x^2+y^2+z^2$$
Equality holds for $x=y=z = 3$.
To maximize $x^2+y^2+z^2$ subject to the constraint $\dfrac1x+\dfrac1y+\dfrac1z=1$, you could use the Lagrange multiplier method and look at $x^2+y^2+z^2-\lambda\left(\dfrac1x+\dfrac1y+\dfrac1z-1\right)$. Take derivatives and see that $2x+\dfrac\lambda{x^2}=2y+\dfrac\lambda{y^2}=2z+\dfrac\lambda{z^2}=\dfrac1x+\dfrac1y+\dfrac1z-1=0,$ so $2x^3+\lambda=2y^3+\lambda=2z^3+\lambda,$ so $x=y=z$ and $x=3$ and $x^2+y^2+z^2=27$.
If you wish to complete your AM-GM, note $$1=\frac1x+\frac1y+\frac1z\geqslant \frac3{\sqrt[3]{xyz}}\implies xyz\geqslant 27$$
Using that with what you got, viz. $$\left(\frac{x^2+y^2+z^2}3 \right)^3 \geqslant (xyz)^2\geqslant 27^2=9^3 \implies x^2+y^2+z^2\geqslant 27$$
Also, the Tangent Line method helps: $$x^2+y^2+z^2-27=\sum_{cyc}(x^2-9)=$$ $$=\sum_{cyc}\left(x^2-9-18\left(\frac{1}{x}-\frac{1}{3}\right)\right)=\sum_{cyc}\frac{(x-3)^2(x+6)}{x}\geq0.$$
If you are trying to use GM<=AM, I suppose you have $x,y,z\geq 0$. Check this link QM>=AM>=GM>=HM
First Step
Using HM<=GM:
$$ \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \leq \sqrt[3]{xyz} \implies xyz\geq \frac{3^3}{1}=27$$
Second Step
Using the AM<=QM:
$$ \frac{x+y+z}{3} \leq \sqrt{\frac{x^2+y^2+z^2}{3}}\implies \frac{x^2+y^2+z^2+2(xy+xz+yz)}{9}\leq \frac{x^2+y^2+z^2}{3}\implies \frac{x^2+y^2+z^2}{3}+\frac{2(xy+xz+yz)}{3}\leq x^2+y^2+z^2 \implies \frac{2}{3}(xy+xz+yz)\leq \frac{2}{3}(x^2+y^2+z^2)\implies xy+xz+yz\leq x^2+y^2+z^2$$
We know that $xy+xz+yz = xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=xyz$. Thus by First Step
$$27\leq xyz \leq x^2+y^2+z^2$$
To finish this proof, note that $(x,y,z)=(3,3,3)$ is a feasible solution with an objective value equal to $27$.
Use lagrange multipliers. $\nabla f=\langle 2x,2y,2z\rangle$ and $\nabla g=\langle x^{-2},y^{-2},z^{-2}\rangle$ where $f(x,y,z)=x^2+y^2+z^2$ and $g(x,y,z)= x^{-1}+y^{-1}+z^{-1}$. Solving $\nabla f=\lambda\nabla g$ yields $\lambda=2x^3=2y^3=2z^3$ and therefore $x=y=z$. Plugging this into $g(x,y,z)=1$ we have $x=y=z=3$ which we plug into $f(x,y,z)$ to obtain $3^2+3^2+3^2=27$ which is obviously a minimum as $x^2+y^2+z^2$ is unbounded above on the surface $g(x,y,z)=1$.
You can still finish it off if you notice that $( xy+yz+zx )^2 = (xyz)^2 = xy\cdot yz \cdot zx \le \dfrac{(xy+yz+zx)^3}{27} \implies xy+yz+zx \ge 27 \implies \dfrac{x^2+y^2+z^2}{3} \ge \sqrt[3]{27^2} = 9 \implies x^2+y^2+z^2 \ge 9\cdot 3 = 27 $.
