MHB Proof of Cauchy's Inequality .... Sohrab Proposition 2.1.23 ....

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Inequality Proof
Click For Summary
The discussion revolves around proving Proposition 2.1.23 from Houshang H. Sohrab's "Basic Real Analysis," specifically focusing on the properties of the discriminant in the context of the quadratic function F(t). Participants clarify that since F(t) is non-negative for all t, the discriminant D must be non-positive, leading to the conclusion that Z^2 is less than or equal to XY. This conclusion is crucial as it supports Cauchy's inequality. The thread highlights the importance of understanding how the discriminant's properties relate to the non-negativity of the quadratic function. Overall, the discussion emphasizes the mathematical principles underlying the proof of the inequality.
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

I need help with Proposition 2.1.23/Exercise 2.1.24 (Exercise 2.1.24 asks readers to prove Proposition 2.1.23) ...

Proposition 2.1.23/Exercise 2.1.24 reads as follows:View attachment 7055
In the above text by Sohrab, we read the following:

" ... ... Observe that for any $$t \in \mathbb{R}, F(t) = \sum_{ i = 1 }^n ( x_i - t y_i )^2 \ge 0$$ and look at the discriminant $$Z^2 - XY$$ of $$F(t)$$. ... ... "Can someone please explain how one proceeds after determining the discriminant and what principles/properties of the discriminant are used?

Help will be appreciated ...

Peter

=========================================================================================Note: my working so far on the exercise is as follows:$$F(t) = \sum_{ i = 1 }^n ( x_i - t y_i )^2 \ge 0$$

Now ... ... $$\sum_{ i = 1 }^n ( x_i - t y_i )^2 \ge 0$$

$$\Longleftrightarrow \sum_{ i = 1 }^n ( x_i^2 - 2t x_i y_i + t^2 y_i^2) \ge 0$$

$$\Longleftrightarrow \sum_{ i = 1 }^n x_i^2 - 2 t \sum_{ i = 1 }^n x_i y_i + t^2 \sum_{ i = 1 }^n y_i^2 \ge 0$$ ... ... ... (1)... now let $$X = \sum_{ i = 1 }^n x_i^2$$, $$Y = \sum_{ i = 1 }^n y_i^2$$ and $$Z = \sum_{ i = 1 }^n x_i y_i$$Then (1) becomes $$X - 2t Z + Y t^2 \ge 0$$

and the discriminant, $$D = 4Z^2 - 4XY$$ BUT ... how do we proceed from here .. ... and exactly what properties of the discriminant do we use ...
 
Physics news on Phys.org
Peter said:
" ... ... Observe that for any $$t \in \mathbb{R}, F(t) = \sum_{ i = 1 }^n ( x_i - t y_i )^2 \ge 0$$ and look at the discriminant $$Z^2 - XY$$ of $$F(t)$$. ... ... "Can someone please explain how one proceeds after determining the discriminant and what principles/properties of the discriminant are used?

Help will be appreciated ...

Peter

=========================================================================================Note: my working so far on the exercise is as follows:$$F(t) = \sum_{ i = 1 }^n ( x_i - t y_i )^2 \ge 0$$

Now ... ... $$\sum_{ i = 1 }^n ( x_i - t y_i )^2 \ge 0$$

$$\Longleftrightarrow \sum_{ i = 1 }^n ( x_i^2 - 2t x_i y_i + t^2 y_i^2) \ge 0$$

$$\Longleftrightarrow \sum_{ i = 1 }^n x_i^2 - 2 t \sum_{ i = 1 }^n x_i y_i + t^2 \sum_{ i = 1 }^n y_i^2 \ge 0$$ ... ... ... (1)... now let $$X = \sum_{ i = 1 }^n x_i^2$$, $$Y = \sum_{ i = 1 }^n y_i^2$$ and $$Z = \sum_{ i = 1 }^n x_i y_i$$Then (1) becomes $$X - 2t Z + Y t^2 \ge 0$$

and the discriminant, $$D = 4Z^2 - 4XY$$ BUT ... how do we proceed from here .. ... and exactly what properties of the discriminant do we use ...
If the discriminant is positive then the quadratic equation $F(t) = 0$ will have two distinct roots $t = \dfrac{2Z \pm\sqrt D}{2Y}$. If $t$ lies anywhere between these roots then $F(t)$ will be negative. But $F(t)$ is never negative. It follows that the discriminant is not positive, in other words $D \leqslant 0$. Thus $4Z^2 - 4XY \leqslant 0$, so that $Z^2 \leqslant XY.$
 
Opalg said:
If the discriminant is positive then the quadratic equation $F(t) = 0$ will have two distinct roots $t = \dfrac{2Z \pm\sqrt D}{2Y}$. If $t$ lies anywhere between these roots then $F(t)$ will be negative. But $F(t)$ is never negative. It follows that the discriminant is not positive, in other words $D \leqslant 0$. Thus $4Z^2 - 4XY \leqslant 0$, so that $Z^2 \leqslant XY.$

Thanks Opalg ... your post was really helpful ...

Peter
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K