# Does $A\geq (\mathrm{tr}(A^{-1}))^{-1}I$ hold for symmetric positive-definite matrix $A$?

Somewhere in my reading, it seems that the following inequality holds for every symmetric positive definite matrix $$A$$, $$A\geq \big(\mathrm{tr}(A^{-1})\big)^{-1}I,$$ where $$I$$ is the identity matrix.

Is this true?

Thank you!

Yes. By a change of orthonormal basis, you may assume that $$A=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$$ and the inequality becomes $$\operatorname{tr}(A^{-1})A\ge I$$ or $$\lambda_i\sum_{k=1}^n\frac{1}{\lambda_k}\ge1$$ for every $$i$$.