# Prove the following inequality .

While I was going through matrix norms, I came across the fact that Operator norm of a matrix is less than equal to Frobenius Norm of the matrix. While trying to prove this fact I came across the inequality:-

If $$\sum_{j=1}^{n}x_{j}^{2} = 1$$ then: $$\sum_{j=1}^{n} a_{j}^{2}\geq\left( \sum_{j=1}^{n}a_{j}x_{j}\right)^{2}$$ ($a_{j}$ and $x_{j}$ are real numbers for $j\in\{1,2,3 \ldots\}$)

I want to know if the inequality is true or not. If true what is the proof. So far I have not been able to find any counter example.

• Cauchy-Schwarz? – velut luna May 18 '16 at 8:24
• Its cauchy-shwarz inequality – avz2611 May 18 '16 at 8:24

$$\left( \sum_{j=1}^{n}a_{j}x_{j}\right)^{2} \leq \left(\sum_{j=1}^{n} a_{j}^{2} \right)\left(\sum_{j=1}^{n} x_{j}^{2} \right) = \sum_{j=1}^{n} a_{j}^{2}$$
in the last step I used $$\sum_{j=1}^{n} x_{j}^{2} =1$$