Nesbitt's inequality - Alchetron, The Free Social Encyclopedia

In mathematics, Nesbitt's inequality is a special case of the Shapiro inequality. It states that for positive real numbers a, b and c we have:

First proof: AM-HM

By the AM-HM inequality on ( a + b ) , ( b + c ) , ( c + a ) ,

( a + b ) + ( a + c ) + ( b + c ) 3 ≥ 3 1 a + b + 1 a + c + 1 b + c .

Clearing denominators yields

( ( a + b ) + ( a + c ) + ( b + c ) ) ( 1 a + b + 1 a + c + 1 b + c ) ≥ 9 ,

from which we obtain

2 a + b + c b + c + 2 a + b + c a + c + 2 a + b + c a + b ≥ 9

by expanding the product and collecting like denominators. This then simplifies directly to the final result.

Second proof: Rearrangement

Suppose a ≥ b ≥ c , we have that

1 b + c ≥ 1 a + c ≥ 1 a + b

define

x → = ( a , b , c ) y → = ( 1 b + c , 1 a + c , 1 a + b )

The scalar product of the two sequences is maximum because of the rearrangement inequality if they are arranged the same way, call y → 1 and y → 2 the vector y → shifted by one and by two, we have:

x → ⋅ y → ≥ x → ⋅ y → 1 x → ⋅ y → ≥ x → ⋅ y → 2

Addition yields Nesbitt's inequality.

Third proof: Hilbert's Seventeenth Problem

The following identity is true for all a , b , c :

a b + c + b a + c + c a + b = 3 2 + 1 2 ( ( a − b ) 2 ( a + c ) ( b + c ) + ( a − c ) 2 ( a + b ) ( b + c ) + ( b − c ) 2 ( a + b ) ( a + c ) )

This clearly proves that the left side is no less than 3 2 for positive a,b and c.

Note: every rational inequality can be solved by transforming it to the appropriate identity, see Hilbert's seventeenth problem.

Fourth proof: Cauchy–Schwarz

Invoking the Cauchy–Schwarz inequality on the vectors ⟨ a + b , b + c , c + a ⟩ , ⟨ 1 a + b , 1 b + c , 1 c + a ⟩ yields

( ( b + c ) + ( a + c ) + ( a + b ) ) ( 1 b + c + 1 a + c + 1 a + b ) ≥ 9 ,

which can be transformed into the final result as we did in the AM-HM proof.

Fifth proof: AM-GM

We first employ a Ravi substitution: let x = a + b , y = b + c , z = c + a . We then apply the AM-GM inequality to the set of six values { x 2 z , z 2 x , y 2 z , z 2 y , x 2 y , y 2 x } to obtain

( x 2 z + z 2 x ) + ( y 2 z + z 2 y ) + ( x 2 y + y 2 x ) 6 ≥ x 2 z ⋅ z 2 x ⋅ y 2 z ⋅ z 2 y ⋅ x 2 y ⋅ y 2 x 6 = x y z .

Dividing by x y z / 6 yields

x + z y + y + z x + x + y z ≥ 6.

Substituting out the x , y , z in favor of a , b , c yields

2 a + b + c b + c + a + b + 2 c a + b + a + 2 b + c c + a ≥ 6 ,

which then simplifies directly 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 n real numbers ( x k ) and any sequence of n positive numbers ( a k ) , ∑ k = 1 n x k 2 a k ≥ ( ∑ k = 1 n x k ) 2 ∑ k = 1 n a k . We use its three-term instance with x -sequence a , b , c and a -sequence a ( b + c ) , b ( c + a ) , c ( a + b ) :

a 2 a ( b + c ) + b 2 b ( c + a ) + c 2 c ( a + b ) ≥ ( a + b + c ) 2 a ( b + c ) + b ( c + a ) + c ( a + b )

By multiplying out all the products on the lesser side and collecting like terms, we obtain

a 2 a ( b + c ) + b 2 b ( c + a ) + c 2 c ( a + b ) ≥ a 2 + b 2 + c 2 + 2 ( a b + b c + c a ) 2 ( a b + b c + c a ) ,

which simplifies to

a b + c + b c + a + c a + b ≥ a 2 + b 2 + c 2 2 ( a b + b c + c a ) + 1.

By the rearrangement inequality, we have a 2 + b 2 + c 2 ≥ a b + b c + c a , so the fraction on the lesser side must be at least 1 2 . Thus,

a b + c + b c + a + c a + b ≥ 3 2 .

References

Nesbitt's inequality Wikipedia

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