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# Prove an if and only if condition for equality in the Cauchy-Schwarz inequality

Recall the Cauchy-Schwarz inequality,

For arbitrary real numbers and we have The claim is then that the equality sign holds if and only if there is a real number such that for each .

Proof. ( ) If for all , then equality clearly holds. Assume then that for at least one . Then, considering the equation and defining, We have, But, since we know (by assumption), we have which is in (since since for at least one and each term in nonnegative, so the sum is strictly positive).
( ) Assume there exists such that for each . Then, . So, ### 2 comments

1. Anonymous says:

The forward implication is actually false. Consider 0 as the a-sequence and 9 as the b-sequence. Then the equality holds but there is no real x which will make ax + b = b = 9 = 0 true.

• nu creation says:

I think it might be a mistake in the book. Because the proof technique in the second implication would still work if there is a y s.t. ak + ybk =0.. this would also imply the Cauchy-Schwartz equality which means there’s something’s wrong with the only if since having such an x isn’t the only way to arrive at the equality. Notice if we change the statement to include the bit about the y, his forward implication is true… If sequence a is all zeros, there exists a y st ak + ybk= 0 namely y=0. So make the iff statement require either there is such an x or there is such a y.