If $A$ is positive definite prove that $(x^TAx)(x^TA^{-1}x) \geq 1$

Question

Let $A$ be an $n\times n$ positive definite matrix and $x$ be a unit vector. Prove that $$(x^TAx)(x^TA^{-1}x)\geq 1$$

My attempt:

Since $A$ is positive definite we have that for every non zero vector $x$, $x^TAx>0$. Also if $A$ is positive definite then so is $A^{-1}$. Thus $x^TA^{-1}x>0$. This would imply that for any unit vector $x$, $(x^TAx)(x^TA^{-1}x)>0$. How do I show it is greater than or equal to $1$?

We also have that $\Vert x\Vert=x^Tx=1$. But I am not sure how to use this fact.

Any help would be appreciated.

Answer 1

One can diagonalise $A$ by an orthogonal matrix. Orthogonal matrices preserve norm one vectors. So, we may assume $A$ is diagonal. Then the inequality reduces to proving $$\sum_{i=1}^n a_ix_i^2\sum_{j=1}^n\frac{x_i^2}{a_j}\ge1$$ under the hypotheses that all $a_i>0$ and $\sum_{i=1}^n x_i^2=1$. I'll leave this new inequality for you to figure out.

Answer 2

The same solution as Lord Shark's but from an alternative view: let $P$ be the unique positive definite square root of $A$. Then the left hand side becomes $\|Px\|^2\|P^{-1}x\|^2$. Apply Cauchy–Bunyakovsky–Schwarz inequality to conclude.