Minimum value of $ f(x,y,z)=\left(x+\frac{1}{y}\right)^2+\left(y+\frac{1}{z}\right)^2+\left(z+\frac{1}{x}\right)^2. $ [duplicate]

If $x>0$, $y>0$, $z>0$ and $x+y+z=6$ then find the minimum value of $$ f(x,y,z)=\left(x+\frac{1}{y}\right)^2+\left(y+\frac{1}{z}\right)^2+\left(z+\frac{1}{x}\right)^2. $$

Thanks in advance.

By C-S $$\sum_{cyc}\left(x+\frac{1}{y}\right)^2=\frac{1}{3}\cdot(1+1+1)\sum_{cyc}\left(x+\frac{1}{y}\right)^2\geq\frac{1}{3}\left(\sum_{cyc}\left(x+\frac{1}{x}\right)\right)^2\geq$$ $$\geq\frac{1}{3}\left(6+\frac{9}{6}\right)^2=\frac{75}{4}.$$ The equality occurs for $x=y=z=2$, which says that $\frac{75}{4}$ is an answer.

C-S it's the following. $$(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)^2\geq(a_1b_1+a_2b_2+...+a_nb_n)^2.$$ In our case $(1^2+1^2+1^2)\left(\left(x+\frac{1}{y}\right)^2+\left(y+\frac{1}{z}\right)^2+\left(z+\frac{1}{x}\right)^2\right)\geq\left(x+\frac{1}{y}+y+\frac{1}{z}+z+\frac{1}{x}\right)^2$ and $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{6}(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq\frac{(1+1+1)^2}{6}=\frac{9}{6}$$

Michael Rozenberg's solution is great, and I think this is the way to go. I just want to point out that it is possible to solve this problem just by observing that there is a great deal of symmetry. What we know is that the available resources must add up to $6$, and since the coefficients in front of each variable are $1$, we do not "penalize" any variable more than the others. The same goes with the cost function (where each variable is presented in the same way as the others). Since this is the case we can actually transform this problem into an one variable problem, thus $v=x=y=z$. We can now write the optimization problem as $$ \underset{v>0}{\text{minimize}}\quad 3(v+1/v)^2\\ \text{subject to}\quad v=2 $$ From this, we can directly conclude that the only point that is acceptable is $v=2$, yielding a cost of $75/4$. This is both the maximum and the minimum point (since it is the only point).