# Thread: Cauchy-Schwarz Inequality

1. ## Cauchy-Schwarz Inequality

If $a_{1}, \ldots, a_{n}$ and $b_{1} \ldots, b_{n}$ are arbitrary real numbers, then $\left(\sum_{k=1}^{n} a_{k}b_{k} \right)^{2} \leq \left(\sum_{k=1}^{n} a_{k}^{2} \right) \left(\sum_{k=1}^{n} b_{k}^{2} \right)$. If some $a_{i} \neq 0$ then equality holds if and only if there is a real $x$ such that $a_{k}x + b_{k} = 0$ for each $k = 1,2, \ldots , n$.

So to basically prove this you do the following: $\sum_{k=1}^{n} (a_{k}x + b_{k})^{2} \geq 0$ for all $x$. It is equaled to $0$ if $x_i = 0$.

And this is equaled to $\sum_{k=1}^{n} a_{k}x^{2} + 2a_{k}x b_{k} + b_{k}^{2} \geq 0$

Then we invoke the discriminant of the quadratic equation $Ax^{2} + 2Bx + C \geq 0$ by making the required substitutions?

2. Originally Posted by heathrowjohnny
If $a_{1}, \ldots, a_{n}$ and $b_{1} \ldots, b_{n}$ are arbitrary real numbers, then $\left(\sum_{k=1}^{n} a_{k}b_{k} \right)^{2} \leq \left(\sum_{k=1}^{n} a_{k}^{2} \right) \left(\sum_{k=1}^{n} b_{k}^{2} \right)$. If some $a_{i} \neq 0$ then equality holds if and only if there is a real $x$ such that $a_{k}x + b_{k} = 0$ for each $k = 1,2, \ldots , n$.

So to basically prove this you do the following: $\sum_{k=1}^{n} (a_{k}x + b_{k})^{2} \geq 0$ for all $x$. It is equaled to $0$ if $x_i = 0$.

And this is equaled to $\sum_{k=1}^{n} a_{k}x^{2} + 2a_{k}x b_{k} + b_{k}^{2} \geq 0$

Then we invoke the discriminant of the quadratic equation $Ax^{2} + 2Bx + C \geq 0$ by making the required substitutions?

have you ever tried searching from google or yahoo? in any field, the cauch-schwarz inequality have the same proof..

correction: $\sum_{k=1}^{n} {a_{k}}^2x^{2} + 2a_{k}x b_{k} + b_{k}^{2} \geq 0$

anyways to continue your proof, yes, you must consider the discriminant, and that the discriminant must be $\leq 0$ (why?) (since $Ax^{2} + 2Bx + C \geq 0$, that is, if it is > 0, the equation has no real root and so the discriminant is < 0. ...)
can you continue now??