Prove that $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq 27$ How can I prove $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq 27$, given that $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9$ and $x+y+z=1$.
I've already tried using that: $\frac{1}{x} +\frac{1}{y} +\frac{1}{z}\geq 9$ But I can't seem to manipulate that to prove the above.

You may also try this :
Apply AM-HM inequality on the set $\{x^2,y^2,z^2\}$ :
$$\frac {x^2+y^2+z^2}{3} \geq \frac {3}{\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}}$$
$$\implies \big(x^2+y^2+z^2\big)\big({\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}}\big) \geq 9$$
$$\implies {\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}} \geq \frac{9}{x^2+y^2+z^2}\,\,\,(♦)$$
Now all that remains is to prove that $$x^2+y^2+z^2 \leq \frac{1}{3} \,\,\,(♣)$$
with the constraint that $$x+y+z=1 $$
I am assuming that you are skilled enough to prove $(♣)$. (You can use Lagrange Multipliers or some other technique.)
Plug $(♣)$ into $(♦)$ to get :
$$ {\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}}\geq \frac{9}{1/3} = 27$$
$$HENCE \,\,\,\, PROVED$$
NOTE : Since $x^2+y^2+z^2$ is in the denominator (♦) , the sign of (♣) flips/reverts , thus proving the desired inequality.
