Consider the following statement and "proof". For a real number x , if x>0 then (x^2+2)/(4x) >0 . Proof: Suppose x>0 but (x^2+2)/(4x) \leq 0 . Then we may multiply both sides of (x^2+2)/(4x) by x without changing the inequality, since x>0 . Hence (x^2+2)/4 \leq 0 . Then (x^2+2) \leq 0 . So x^2 \leq -2 . But x>0 implies x^2>0 . This contradiction proves our original assumption was false, so the statement is true. Which of the following are true?

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Consider the following statement and "proof".

For a real number x, if x>0 then (x^2+2)/(4x) >0 .

Proof:

Suppose x>0 but $(x^2+2)/(4x) \leq 0$ (x^2+2)/(4x) \leq 0.
Then we may multiply both sides of (x^2+2)/(4x) by x without changing the inequality, since x>0.
Hence $(x^2+2)/4 \leq 0$ (x^2+2)/4 \leq 0.
Then $(x^2+2) \leq 0$ (x^2+2) \leq 0 .
So $x^2 \leq -2$ x^2 \leq -2 .
But x>0 implies x^2>0.
This contradiction proves our original assumption was false, so the statement is true.