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.

Cauchy-Schwarz inequality tells us that
\begin{equation}
\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2 \leq 3\times \left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\qquad (\star)
\end{equation}
But the left-hand term is $\geq 9^2$, so
$$
\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \geq \frac{9^2}{3} = 27.
$$

Note that in this particular case, it very is easy to prove $(\star)$. Let
$$
A = 3 \left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right),\qquad B = \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2.
$$
We need to show that $A \geq B$. Taking their difference,
\begin{align}
A - B & = \frac{1}{x^2}+\frac{1}{y^2} - 2\frac{1}{xy} + \frac{1}{y^2} + \frac{1}{z^2} - 2\frac{1}{yz} + \frac{1}{z^2}+\frac{1}{x^2} - 2\frac{1}{zx}\\
& = \left(\frac{1}{x}-\frac{1}{y}\right)^2 + \left(\frac{1}{y}-\frac{1}{z}\right)^2 + \left(\frac{1}{z}-\frac{1}{x}\right)^2\\
& \geq 0.
\end{align}
