How to prove $\sum_{i=1}^{k}x_i^2\ge k$ if $\sum_{i=1}^{k}\frac{1}{x_i}=k$ and $\min (x_i-1)^2$ is sufficiently small?

Question

How to prove $\sum_{i=1}^{k}x_i^2\ge k$ if $\sum_{i=1}^{k}\frac{1}{x_i}=k$ and $\min (x_i-1)^2$ is sufficiently small?

In $k=2$ case it is true.

$\frac{1}{x}+\frac{1}{y}=2$ implies $y = \frac{x}{2x-1}$with plot:

Answer 1

Maybe $(x_i-1)^2$ is sufficiently small says that $x_i>0$.

If so, for $x_i>0$ it's true by Holder: $$\sum_{i=1}^kx_i^2=\frac{\sum\limits_{i=1}^kx_i^2\left(\sum\limits_{i=1}^k\frac{1}{x_i}\right)^2}{k^2}\geq\frac{k^3}{k^2}=k.$$ Also, the Tangent Line method helps: $$\sum_{i=1}^kx_i^2-k=\sum_{i=1}^k\left(x_i-1\right)=\sum_{i=1}^k\left(x_i-1+2\left(\frac{1}{x_i}-1\right)\right)=\sum_{i=1}^k\frac{(x_i-1)^2(x_i+2)}{x_i}\geq0.$$