In mathematics, Nesbitt's inequality states that for positive real numbers a, b and c, It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, and was published at least 50 years earlier. There is no corresponding upper bound as any of the 3 fractions in the inequality can be made arbitrarily large. Proof First proof: AM-HM inequality By the AM-HM inequality on , Clearing denominators yields from which we obtain by expanding the product and collecting like denominators. This then simplifies directly to the final result. Second proof: Rearrangement Suppose , we have that define The scalar product of the two sequences is maximum because of the rearrangement inequality if they are arranged the same way, call and the vector shifted by one and by two, we have: Addition yields our desired Nesbitt's inequality. Third proof: Sum of Squares The following identity is true for all This clearly proves that the left side is no less than for positive a, b and c. Note: every rational inequality can be demonstrated by transforming it to the appropriate sum-of-squares identity, see Hilbert's seventeenth problem. Fourth proof: Cauchy–Schwarz Invoking the Cauchy–Schwarz inequality on the vectors yields which can be transformed into the final result as we did in the AM-HM proof. Fifth proof: AM-GM Let . We then apply the AM-GM inequality to obtain the following because Substituting out the in favor of yields which then simplifies to the final result. Sixth proof: Titu's lemma Titu's lemma, a direct consequence of the Cauchy–Schwarz inequality, states that for any sequence of real numbers and any sequence of positive numbers , . We use the lemma on and . This gives, This results in, i.e., Seventh proof: Using homogeneity As the left side of the inequality is homogeneous, we may assume . Now define , , and . The desired inequality turns into , or, equivalently, . This is clearly true by Titu's Lemma. Eighth proof: Jensen inequality Define and consider the function . This function can be shown to be convex in and, invoking Jensen inequality, we get A straightforward computation yields Ninth proof: Reduction to a two-variable inequality By clearing denominators, It now suffices to prove that for , as summing this three times for and completes the proof. As we are done. References Ion Ionescu, Romanian Mathematical Gazette, Volume XXXII (September 15, 1926 - August