First proof: AM-HM inequality
edit
By the AM -HM inequality on
(
a
+
b
)
,
(
b
+
c
)
,
(
c
+
a
)
{\displaystyle (a+b),(b+c),(c+a)}
(
a
+
b
)
+
(
a
+
c
)
+
(
b
+
c
)
3
≥
3
1
a
+
b
+
1
a
+
c
+
1
b
+
c
.
{\displaystyle {\frac {(a+b)+(a+c)+(b+c)}{3}}\geq {\frac {3}{\displaystyle {\frac {1}{a+b}}+{\frac {1}{a+c}}+{\frac {1}{b+c}}}}.}
Clearing denominators yields
(
(
a
+
b
)
+
(
a
+
c
)
+
(
b
+
c
)
)
(
1
a
+
b
+
1
a
+
c
+
1
b
+
c
)
≥
9
,
{\displaystyle ((a+b)+(a+c)+(b+c))\left({\frac {1}{a+b}}+{\frac {1}{a+c}}+{\frac {1}{b+c}}\right)\geq 9,}
from which we obtain
2
a
+
b
+
c
b
+
c
+
2
a
+
b
+
c
a
+
c
+
2
a
+
b
+
c
a
+
b
≥
9
{\displaystyle 2{\frac {a+b+c}{b+c}}+2{\frac {a+b+c}{a+c}}+2{\frac {a+b+c}{a+b}}\geq 9}
by expanding the product and collecting like denominators. This then simplifies directly to the final result.
Second proof: Rearrangement
edit
Suppose
a
≥
b
≥
c
{\displaystyle a\geq b\geq c}
1
b
+
c
≥
1
a
+
c
≥
1
a
+
b
.
{\displaystyle {\frac {1}{b+c}}\geq {\frac {1}{a+c}}\geq {\frac {1}{a+b}}.}
define
x
→
=
(
a
,
b
,
c
)
{\displaystyle {\vec {x}}=(a,b,c)}
and
y
→
=
(
1
b
+
c
,
1
a
+
c
,
1
a
+
b
)
{\displaystyle {\vec {y}}=\left({\frac {1}{b+c}},{\frac {1}{a+c}},{\frac {1}{a+b}}\right)}
By the rearrangement inequality, the dot product of the two sequences is maximized when the terms are arranged to be both increasing or both decreasing. The order here is both decreasing. Let
y
→
1
{\displaystyle {\vec {y}}_{1}}
y
→
2
{\displaystyle {\vec {y}}_{2}}
y
→
{\displaystyle {\vec {y}}}
x
→
⋅
y
→
≥
x
→
⋅
y
→
1
{\displaystyle {\vec {x}}\cdot {\vec {y}}\geq {\vec {x}}\cdot {\vec {y}}_{1}}
x
→
⋅
y
→
≥
x
→
⋅
y
→
2
{\displaystyle {\vec {x}}\cdot {\vec {y}}\geq {\vec {x}}\cdot {\vec {y}}_{2}}
Addition then yields Nesbitt's inequality.
Third proof: Sum of Squares
edit
The following identity is true for all
a
,
b
,
c
:
{\displaystyle 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
)
)
.
{\displaystyle {\frac {a}{b+c}}+{\frac {b}{a+c}}+{\frac {c}{a+b}}={\frac {3}{2}}+{\frac {1}{2}}\left({\frac {(a-b)^{2}}{(a+c)(b+c)}}+{\frac {(a-c)^{2}}{(a+b)(b+c)}}+{\frac {(b-c)^{2}}{(a+b)(a+c)}}\right).}
This clearly proves that the left side is no less than
3
2
{\displaystyle {\frac {3}{2}}}
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
edit
Invoking the Cauchy–Schwarz inequality on the vectors
⟨
a
+
b
,
b
+
c
,
c
+
a
⟩
,
⟨
1
a
+
b
,
1
b
+
c
,
1
c
+
a
⟩
{\displaystyle \displaystyle \left\langle {\sqrt {a+b}},{\sqrt {b+c}},{\sqrt {c+a}}\right\rangle ,\left\langle {\frac {1}{\sqrt {a+b}}},{\frac {1}{\sqrt {b+c}}},{\frac {1}{\sqrt {c+a}}}\right\rangle }
(
(
b
+
c
)
+
(
a
+
c
)
+
(
a
+
b
)
)
(
1
b
+
c
+
1
a
+
c
+
1
a
+
b
)
≥
9
,
{\displaystyle ((b+c)+(a+c)+(a+b))\left({\frac {1}{b+c}}+{\frac {1}{a+c}}+{\frac {1}{a+b}}\right)\geq 9,}
which can be transformed into the final result as we did in the AM-HM proof .
Fifth proof: AM-GM
edit
Let
x
=
a
+
b
,
y
=
b
+
c
,
z
=
c
+
a
{\displaystyle x=a+b,y=b+c,z=c+a}
AM-GM inequality to obtain
x
+
z
y
+
y
+
z
x
+
x
+
y
z
≥
6
,
{\displaystyle {\frac {x+z}{y}}+{\frac {y+z}{x}}+{\frac {x+y}{z}}\geq 6,}
because
x
y
+
z
y
+
y
x
+
z
x
+
x
z
+
y
z
≥
6
x
y
⋅
z
y
⋅
y
x
⋅
z
x
⋅
x
z
⋅
y
z
6
=
6.
{\displaystyle {\frac {x}{y}}+{\frac {z}{y}}+{\frac {y}{x}}+{\frac {z}{x}}+{\frac {x}{z}}+{\frac {y}{z}}\geq 6{\sqrt[{6}]{{\frac {x}{y}}\cdot {\frac {z}{y}}\cdot {\frac {y}{x}}\cdot {\frac {z}{x}}\cdot {\frac {x}{z}}\cdot {\frac {y}{z}}}}=6.}
Substituting out the
x
,
y
,
z
{\displaystyle x,y,z}
a
,
b
,
c
{\displaystyle a,b,c}
2
a
+
b
+
c
b
+
c
+
a
+
b
+
2
c
a
+
b
+
a
+
2
b
+
c
c
+
a
≥
6
{\displaystyle {\frac {2a+b+c}{b+c}}+{\frac {a+b+2c}{a+b}}+{\frac {a+2b+c}{c+a}}\geq 6}
2
a
b
+
c
+
2
c
a
+
b
+
2
b
a
+
c
+
3
≥
6
,
{\displaystyle {\frac {2a}{b+c}}+{\frac {2c}{a+b}}+{\frac {2b}{a+c}}+3\geq 6,}
which then simplifies to the final result.
Sixth proof: Titu's lemma
edit
Titu's lemma , a direct consequence of the Cauchy–Schwarz inequality , states that for any sequence of
n
{\displaystyle n}
(
x
k
)
{\displaystyle (x_{k})}
n
{\displaystyle n}
(
a
k
)
{\displaystyle (a_{k})}
∑
k
=
1
n
x
k
2
a
k
≥
(
∑
k
=
1
n
x
k
)
2
∑
k
=
1
n
a
k
.
{\displaystyle \displaystyle \sum _{k=1}^{n}{\frac {x_{k}^{2}}{a_{k}}}\geq {\frac {(\sum _{k=1}^{n}x_{k})^{2}}{\sum _{k=1}^{n}a_{k}}}.}
We use the lemma on
(
x
k
)
=
(
1
,
1
,
1
)
{\displaystyle (x_{k})=(1,1,1)}
(
a
k
)
=
(
b
+
c
,
a
+
c
,
a
+
b
)
{\displaystyle (a_{k})=(b+c,a+c,a+b)}
1
b
+
c
+
1
c
+
a
+
1
a
+
b
≥
3
2
2
(
a
+
b
+
c
)
,
{\displaystyle {\frac {1}{b+c}}+{\frac {1}{c+a}}+{\frac {1}{a+b}}\geq {\frac {3^{2}}{2(a+b+c)}},}
which results in
a
+
b
+
c
b
+
c
+
a
+
b
+
c
c
+
a
+
a
+
b
+
c
a
+
b
≥
9
2
{\displaystyle {\frac {a+b+c}{b+c}}+{\frac {a+b+c}{c+a}}+{\frac {a+b+c}{a+b}}\geq {\frac {9}{2}}}
a
b
+
c
+
b
c
+
a
+
c
a
+
b
≥
9
2
−
3
=
3
2
.
{\displaystyle {\frac {a}{b+c}}+{\frac {b}{c+a}}+{\frac {c}{a+b}}\geq {\frac {9}{2}}-3={\frac {3}{2}}.}
Seventh proof: Using homogeneity
edit
As the left side of the inequality is homogeneous, we may assume
a
+
b
+
c
=
1
{\displaystyle a+b+c=1}
x
=
a
+
b
{\displaystyle x=a+b}
y
=
b
+
c
{\displaystyle y=b+c}
z
=
c
+
a
{\displaystyle z=c+a}
1
−
x
x
+
1
−
y
y
+
1
−
z
z
≥
3
2
{\displaystyle {\frac {1-x}{x}}+{\frac {1-y}{y}}+{\frac {1-z}{z}}\geq {\frac {3}{2}}}
1
x
+
1
y
+
1
z
≥
9
/
2
{\displaystyle {\frac {1}{x}}+{\frac {1}{y}}+{\frac {1}{z}}\geq 9/2}
Eighth proof: Jensen's inequality
edit
Let
S
=
a
+
b
+
c
{\displaystyle S=a+b+c}
f
(
x
)
=
x
S
−
x
{\displaystyle f(x)={\frac {x}{S-x}}}
[
0
,
S
]
{\displaystyle [0,S]}
Jensen's inequality , we get
a
S
−
a
+
b
S
−
b
+
c
S
−
c
3
≥
S
/
3
S
−
S
/
3
.
{\displaystyle \displaystyle {\frac {{\frac {a}{S-a}}+{\frac {b}{S-b}}+{\frac {c}{S-c}}}{3}}\geq {\frac {S/3}{S-S/3}}.}
A straightforward computation then yields
a
b
+
c
+
b
c
+
a
+
c
a
+
b
≥
3
2
.
{\displaystyle {\frac {a}{b+c}}+{\frac {b}{c+a}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}.}
Ninth proof: Reduction to a two-variable inequality
edit
By clearing denominators,
a
b
+
c
+
b
a
+
c
+
c
a
+
b
≥
3
2
⟺
2
(
a
3
+
b
3
+
c
3
)
≥
a
b
2
+
a
2
b
+
a
c
2
+
a
2
c
+
b
c
2
+
b
2
c
.
{\displaystyle {\frac {a}{b+c}}+{\frac {b}{a+c}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}\iff 2(a^{3}+b^{3}+c^{3})\geq ab^{2}+a^{2}b+ac^{2}+a^{2}c+bc^{2}+b^{2}c.}
It therefore suffices to prove that
x
3
+
y
3
≥
x
y
2
+
x
2
y
{\displaystyle x^{3}+y^{3}\geq xy^{2}+x^{2}y}
(
x
,
y
)
∈
R
+
2
{\displaystyle (x,y)\in \mathbb {R} _{+}^{2}}
(
x
,
y
)
=
(
a
,
b
)
,
(
a
,
c
)
,
{\displaystyle (x,y)=(a,b),\ (a,c),}
(
b
,
c
)
{\displaystyle (b,c)}
As
x
3
+
y
3
≥
x
y
2
+
x
2
y
⟺
(
x
−
y
)
(
x
2
−
y
2
)
≥
0
{\displaystyle x^{3}+y^{3}\geq xy^{2}+x^{2}y\iff (x-y)(x^{2}-y^{2})\geq 0}
![{\displaystyle {\frac {x}{y}}+{\frac {z}{y}}+{\frac {y}{x}}+{\frac {z}{x}}+{\frac {x}{z}}+{\frac {y}{z}}\geq 6{\sqrt[{6}]{{\frac {x}{y}}\cdot {\frac {z}{y}}\cdot {\frac {y}{x}}\cdot {\frac {z}{x}}\cdot {\frac {x}{z}}\cdot {\frac {y}{z}}}}=6.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1507dc4a3297ab152b5f0dcf02fe8aa41ef56045)
![{\displaystyle [0,S]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ffb2ba8aa6b9d8b265134c1a9857440d4b29d173)
