Prove that
Condition: a, b, c, d, e, f are positive real numbers.
Edit: I'm not sure how awake I was when I typed this. This is to remain here purely to remind me of my own stupidity.
Take the first part:
Add the two by multiplying both with the denominator of the other.
bc occurs both in the numerator and denominator, so this simplifies to
You can do this with the third and forth as well as the fifth and sixth fractions yielding an expression in the form of: and because all the numbers are positive and real, this fraction will never produce a negative number.
Who told you?
Anyway, the inequality is the Shapiro inequality for . You might find a proof along the same lines as those given for the simpler Nesbitt’s inequality on the Wikipedia page.