## Optional exercise (+5 points) A generalization ...

Strengthening the inequality

I’m working through strengthening the Cauchy-Schwarz inequality for real symmetric matrices A and M, aiming to bound the absolute value of the bilinear form <M x, y> by the product of the norms of x and y.

Breaking down the inequality

Taking a closer look at expanding <M (x + t y), x + t y> using the bilinear form and quadratic form, while keeping in mind the symmetry of matrix M and ensuring the argument holds for all t in R.

Refining the expansion

Interestingly enough, I’m refining the expansion of |<M (x + t y), (x + t y)>| and <A (x + t y), (x + t y)> by considering the symmetry of matrix M for all t in R.

Mapping out steps

I’m examining the function f(t) = <M x, x> + 2 t < M x,y> + t^2 <M y,y> with the constraint |f(t)| <= g(t) = <A x, x> + 2 t <A x,y> + t^2 <A y,y> to prove the inequality.

Evaluating options

Mapping out polynomial methods and considering the discriminant's positivity or maximum in t to derive an inequality. Progress is being made by examining key components and approaches.

Raising polynomial insights

OK, let me see. By setting F(t) = f(t)^2 - g(t)^2, I'm evaluating the constraints and ensuring all elements align to form the desired inequality.

Analyzing polynomial constraints

OK, let's break it down: by examining the polynomial f(t)^2 - g(t)^2, I identify constraints on <M x,y>, notably focusing on the coefficient of t^2.

Assessing constraints

I conclude that the leading term <M y,y>^2 - <A y,y>^2 is less than or equal to zero, given the inequality |<M x,x>| <= <A x,x>.

Confirming polynomial negativity

If <M y,y>^2 < <A y,y>^2, the leading coefficient is negative, making the polynomial negative for large t, which aligns with the inequality |f(t)| <= g(t).

Revisiting Cauchy-Schwarz applications

I’m thinking through how applying the Cauchy-Schwarz inequality with different parameters can dramatically simplify the polynomial analysis. This tactic streamlines proving the polynomial's non-negativity.

Examining absolute value constraints

I'm looking at the absolute value conditions on both sides, examining the polynomial terms to ensure they align with the inequality constraints. Progressing through each term carefully.

Reworking polynomial bounds

I’m reworking the inequalities, defining polynomial L(t) and U(t), to simplify the analysis of the polynomial's terms and ensure they align with the given constraints.

Analyzing quadratic terms

I'm working through the polynomial terms |M x,y| and aligning them with < A x,y> to establish bounds and inequalities. This leads to a clearer understanding of the polynomial's behavior.

Clarifying relationships

I’m showing that |< M x,y>| is bounded by expressions involving A, B, C, and D by leveraging standard absolute value properties, leading to a squared inequality for verification.

Tracing absolute value limits

Interestingly enough, by setting constraints on A and B, I’m ensuring their combined limit adheres to $(C+D)/4$, simplifying the inequality for verification.

Revisiting inequality approach

OK, let me see. Using arithmetic-geometric mean inequality, I’m ensuring $(1/2)[<A x,x> + <A y,y>]$ stays above $\sqrt{<A x,x><A y,y>}$, guaranteeing the polynomial inequality holds.

Reworking bounds

Hmm, I’m thinking about adjusting the Cauchy-Schwarz approach to fit the polynomials, given the constraints on $|< M x,x>|$ and $|< A x,x>|$. Progress is being made!

Defining functions

I'm setting up $f(t)$ and $g(t)$ to link polynomial dot products with constraints. These functions will guide the proof alongside Cauchy-Schwarz.

Linking polynomial expansions

I’m digging into polynomial dot products, connecting $< M x,y>$ to $f(t) + i g(t)$ and applying a special identity to sharpen the inequality.

Next steps

Restricting $f(t)$ and $g(t)$ bounds to ensure $< M x,y>^2 \leq < A x,x> < A y,y>$ through the Cauchy-Schwarz inequality's electrical approach.

Crafting a refinement

I’m focusing on a standard approach to show $|< M (x + \alpha y), x + \alpha y >| \leq < A (x + \alpha y), x + \alpha y >$ by leveraging the given assumption.

Formulating constraints

I'm piecing together a polynomial to show $\text{F(alpha)} \geq 0$ for all $\alpha$. This involves expanding dot products and ensuring the polynomial's discriminant remains negative.

Testing polynomial viability

I’ve defined $G(\alpha)$ as a polynomial of degree 4. By comparing the squares of dot products, I ensure $G(\alpha) \geq 0$ for all $\alpha$.

Pulling together constraints

I’m pulling together polynomial constraints, showing the leading coefficient $s^2 - w^2 \geq 0$. This implies $< M y,y> \leq < A y,y >$.

Establishing connections

I'm noting that $< A y,y> \geq | < M y,y> |$ and $< A x,y>^2 \geq < M x,y>^2$ which are derived from the constraints. Now, I’ll explore the discriminant of a quartic polynomial.

Revisiting the approach

I’m rethinking the polynomial $H(X)$ by using $\alpha^2$ and ensuring $\alpha$ is nonnegative. This leads me to consider $f(t) = < M(x + t y), x + t y > + < M(x - t y), x - t y >$ to establish nonnegativity.

Adjusting my approach

I’m improving the $f(t)$ function by adding and subtracting terms to ensure it's non-negative for all $t$. This involves examining inequalities and simplifying the expression.

Establishing inequalities

I’m incorporating terms to show that $a(t)$ must be non-negative, leading to the inequality $|< M x,x> + t^2 < M y,y>| + |< M x,x> - t^2 < M y,y>| \leq 2 < A x,x> + 2 t^2 < A y,y>$.

Applying the identity

By using the identity for $|p+q| + |p-q|$, I’m simplifying $|< M x,x> + t^2 < M y,y>| + |< M x,x> - t^2 < M y,y>|$ to $2 \max(|< M x,x> + t^2 < M y,y>|, |< M x,x> - t^2 < M y,y>|)$.

Reassessing the inequality

OK, let's consider an alternative approach: setting $p = a(t)$ and $q = 2 t < M x,y>$. This leads to a simpler inequality: $|a(t) + 2 t < M x,y>| + |a(t) - 2 t < M x,y>| \leq 2 [|a(t)| + 2 |t| |< M x,y>| ]$.

Reaching the conclusion

I derived an inequality: $|< M x,x> + 2 t < M x,y> + t^2 < M y,y>| + |< M x,x> - 2 t < M x,y> + t^2 < M y,y>| \leq 2 < A x,x> + 2 t^2 < A y,y>$.