I am trying to show that if
$||E(\alpha)|| <=||\alpha||$ -($1$) where $E$ is any projection operator then ,
$$ 2Re(\gamma-\alpha|\gamma-E(\alpha))>||\gamma-\alpha||^2$$.....$(2)$
This is a very common question in stackexcahnge (if ($1$) holds then $E$ is an orthogonal projection) but I was trying to do it in a different way .I started with some back calculation ,
My method was to prove that $||\alpha-E(\alpha)||<=||\alpha-\gamma||$.So I assumed this to be true then ,
$$||\alpha-E(\alpha)||^2<=||\alpha-\gamma||^2$$
Now , we take the vector $$||\alpha-\gamma + \gamma -E(\alpha)||^2=||\alpha-\gamma||^2+||\gamma -E(\alpha)||^2+2RE(<\alpha-\gamma|\gamma-E(\alpha)>)$$,
Now,$$||\alpha-\gamma + \gamma -E(\alpha)||^2=||\alpha-\gamma||^2+||\gamma -E(\alpha)||^2-2RE(<\gamma-\alpha|\gamma-E(\alpha)>)$$,
$$||\alpha -E(\alpha)||^2-||\alpha-\gamma||^2=||\gamma -E(\alpha)||^2+2RE(<\alpha-\gamma|\gamma-E(\alpha)>)$$,
So from here I want to conclude the equation,$(2)$
Does $2$ hold true if not can someone give me contradictory example.