TheoremBase

The classical argument: expand 0 <= |x - t y|^2 with t = <x,y>/|y|^2 and compare squares.

Proof

Let x,y∈Ex,y\in E. We use the notation and claims of Elementary Identities in a Real Inner Product Space.

If y=0Ey=0_{E}, then ⟨x,y⟩=0\langle x,y\rangle=0 and ∣y∣=0|y|=0 by Elementary Identities in a Real Inner Product Space (ii), so both sides of the asserted inequality equal 00 (using ∣0∣=0|0|=0, claim 1 of Properties of the Absolute Value in an Ordered Field), and there is nothing to prove.

Suppose yβ‰ 0Ey\ne 0_{E}. Then 0<∣y∣0<|y| by Elementary Identities in a Real Inner Product Space (iii), hence 0<∣y∣20<|y|^{2} by claim 5 of Elementary Order Arithmetic in an Ordered Field, and ∣y∣2|y|^{2} has a multiplicative inverse. Put t=⟨x,yβŸ©β€‰(∣y∣2)βˆ’1∈Rt=\langle x,y\rangle\,(|y|^{2})^{-1}\in\mathbb{R}. By Elementary Identities in a Real Inner Product Space (v), Elementary Identities in a Real Inner Product Space (i) and Elementary Identities in a Real Inner Product Space (iv),

0β‰€βˆ£xβˆ’ty∣2=∣x∣2βˆ’2⟨x,ty⟩+∣ty∣2=∣x∣2βˆ’2t⟨x,y⟩+t2∣y∣2,0\le|x-ty|^{2}=|x|^{2}-2\langle x,ty\rangle+|ty|^{2}=|x|^{2}-2t\langle x,y\rangle+t^{2}|y|^{2},

the first inequality being condition (d) of Real Inner Product Spaces (i), since ∣xβˆ’ty∣2=⟨xβˆ’ty,xβˆ’ty⟩|x-ty|^{2}=\langle x-ty,x-ty\rangle. With the chosen tt we have t⟨x,y⟩=⟨x,y⟩2(∣y∣2)βˆ’1t\langle x,y\rangle=\langle x,y\rangle^{2}(|y|^{2})^{-1} and t2∣y∣2=⟨x,y⟩2(∣y∣2)βˆ’1t^{2}|y|^{2}=\langle x,y\rangle^{2}(|y|^{2})^{-1}, so the display reads

0β‰€βˆ£x∣2βˆ’βŸ¨x,y⟩2 (∣y∣2)βˆ’1.0\le|x|^{2}-\langle x,y\rangle^{2}\,(|y|^{2})^{-1}.

Multiplying by the positive number ∣y∣2|y|^{2} (claim 5 of Elementary Arithmetic in an Ordered Field) and rearranging by the translation rule (claim 3 of Elementary Arithmetic in an Ordered Field) gives ⟨x,y⟩2β‰€βˆ£x∣2∣y∣2=(∣xβˆ£β€‰βˆ£y∣)2\langle x,y\rangle^{2}\le|x|^{2}|y|^{2}=(|x|\,|y|)^{2}. Now ∣⟨x,y⟩∣2=⟨x,y⟩2|\langle x,y\rangle|^{2}=\langle x,y\rangle^{2} because ∣⟨x,y⟩∣|\langle x,y\rangle| equals ⟨x,y⟩\langle x,y\rangle or βˆ’βŸ¨x,y⟩-\langle x,y\rangle (claim 1 of Properties of the Absolute Value in an Ordered Field), and both ∣⟨x,y⟩∣|\langle x,y\rangle| and ∣xβˆ£β€‰βˆ£y∣|x|\,|y| are nonnegative (the same claim; and 0=∣xβˆ£β‹…0β‰€βˆ£xβˆ£β€‰βˆ£y∣0=|x|\cdot 0\le|x|\,|y| by claim 5 of Elementary Arithmetic in an Ordered Field). Therefore ∣⟨x,yβŸ©βˆ£β‰€βˆ£xβˆ£β€‰βˆ£y∣|\langle x,y\rangle|\le|x|\,|y| by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field.

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