Original: > 12 + cb 4 .

Original: Hence
Revised: Hence

Original: [Equation]
Revised: [Equation]

Original: Case 

Revised: Case

Original: If 3 b 3 4 − ba 2 − 23 a 3 ≤ 0 , then
Revised: If 3 b 3 4 − ba 2 − 23 a 3 ≤ 0 , then

Original: [Equation]
Revised: [Equation]

Original: Case 

Revised: Case

Original: If 3 b 3 4 − ba 2 − 23 a 3 > 0 , then
Revised: If 3 b 3 4 − ba 2 − 23 a 3 > 0 , then

Original: [Equation]
Revised: [Equation]

Original: Using Lemma 58, a > b 2 , ( b − a ) 2 ( b + a )
Revised: Using Lemma 58, a > b 2 , ( b − a ) 2 ( b + [Equation]

Original: = ( b 2 − 
Revised: = [Equation] b 2 − [Equation]

Original: [Equation]
Revised: [Equation]

Original: 2 )( b − a ) ≤ b 2 ( b − a ) ≤ b 

Revised: b − [Equation] ) ≤ b 2 ( b − a ) ≤ b

Original: Lemma 

Revised: Lemma

Original: When 0 ≤ a ≤ b, 0 ≤ c ≤ b − a b 2 , b 2 ≥ a (cid:113) a 2 +2 b 2 2(1 − cb ) ≥ a 2 + b 2 2 − ca − cb , we have
Revised: When 0 ≤ a ≤ b, 0 ≤ c ≤ b − a b 2 , b 2 ≥ a (cid:113) a 2 +2 b 2 2(1 − cb ) ≥ a 2 + b 2 2 − ca − cb , we have

Original: [Equation]
Revised: [Equation]

Original: Proof
Revised: Proof

Original: Define
Revised: Define

Original: [Equation]
Revised: [Equation]

Original: := sup S a ( c, b ) ≤ b − cb
Revised: , b ) : = sup S a ( c, b ) ≤ b − cb

Original: Here we suppose WLOG S a ( c, b ) ̸ = ϕ .
Revised: Here we suppose WLOG S a ( c, b ) ̸ = ϕ .

Original: Consider dF( a
Revised: Consider dF

Original: [Equation]
Revised: [Equation]

Original: Define u :=
Revised: Define u :=

Original: cb, v :=
Revised: cb, v :=

Original: ab , then u + v ≤ 1 .
Revised: ab , then u + v ≤ 1 .

Original: [Equation]
Revised: [Equation]

Original: As (cid:113)
Revised: As (cid:113)

Original: [Equation]
Revised: [Equation]

Original: ≥ (1 − u ) ,we have
Revised: ≥ ( [Equation] u ) ,we have

Original: [Equation]
Revised: [Equation]

Original: The above inequality shows that F ( a ) is convex w.r.t to a for a min ( c, b ) ≤ a ≤ a max ( c, b ) .
Revised: The above inequality shows that F ( a ) is convex w.r.t to a for a min ( c, b ) ≤ a ≤ a max ( c, b ) .

Original: Hence F ( a ) ≤ max ( F ( a min ( c, b )) , F ( a max ( c, b )))
Revised: Hence F ( a ) ≤ max ( F ( a min ( c, b )) , F ( a max ( c, b )))

Original: Part 1 We abuse the notation and use a min a shorthand for a min ( c, b ) .
Revised: Part 1 We abuse the notation and use a min a shorthand for a min ( c, b ) .

Original: We have a min (cid:113) a 2 min +2 b 2 2(1 − cb )
Revised: We have a min (cid:113) a 2 min +2 b 2 2(1 − cb )

Original: =
Revised: =

Original: a 2 min + b 2 2 − ca min − cb .
Revised: a 2 min + b 2 2 − ca min − cb .

Original: This implies 2 a min (cid:114)
Revised: This implies 2 a min (cid:114)

Original: [Equation]
Revised: [Equation]

Original: Hence using Lemma 43,
Revised: Hence using Lemma 43,

Original: [Equation]
Revised: [Equation]

Original: Part
Revised: Part

Original: We abuse the notation and use a max a shorthand for a
Revised: We abuse the notation and use a max a shorthand for a

Original: max  c, b 
Revised: max ( c, b )

Original: It’s not easy to see which boundary condition a max satisfy, hence we will discuss bycases.
Revised: It’s not easy to see which boundary condition a max satisfy, hence we will discuss by

Original:  Case 1 a max (cid:113)a 2 max +2 b
Revised: cases Case 1 a max (cid:113)

Original: [Equation]
Revised: [Equation]

Original: 1 − cb
Revised: +

Original: [Equation]
Revised: [Equation]

Original: = a 2 max + b 2 2 − ca max −
Revised: max [Equation]

Original:  cb , in this case we simply redo the calculation in Part 1.
Revised: 2 − ca max − cb , in this case we simply redo the calculation in Part 1.

Original: Case 2 b 2 = a max (cid:113) a 2 max +2 b 2 2(1 − cb ) 
Revised: Case 2 b 2 = a max (cid:113) a 2 max +2 b 2 2(1 − cb ) .

Original: This implies
Revised: This implies

Original: [Equation]
Revised: [Equation]

Original: Hence using Lemma60,
Revised: Hence using Lemma

Original: [Equation]
Revised: [Equation]

Original: Case
Revised: Case

Original: [Equation]
Revised: [Equation]

Original: and b 2 ≥ a max (cid:113) b ( a 2 max +2 b 2 ) 2 a max .
Revised: and b 2 ≥ a max (cid:113) b ( a 2 max +2 b 2 ) 2 a max .

Original: This implies a 3 max + 2 a max b 2 − 2 b 3 ≤ 0 ⇒ a max < 910 b .
Revised: This implies a 3 max + 2 a max b 2 − 2 b 3 ≤ 0 ⇒ a max < 910 b .

Original: Define v :=
Revised: Define v :=

Original: a max b , cb = 1 − v
Revised: a max b , cb = 1 − v

Original: Then by Lemma 58, 12 ≤ v ≤
Revised: Then by Lemma 58, 12 ≤ v ≤

Original: [Equation]
Revised: [Equation]

Original: We will prove the following inequality,
Revised: We will prove the following inequality,

Original: [Equation]
Revised: [Equation]

Original: In fact we can directly showv
Revised: In fact we can directly show

Original: [Equation]
Revised: [Equation]

Original: for v ∈ [0 . 5 , 0 . 9] .
Revised: for v ∈ [0 . 5 , 0 . 9] .

Original: Hence we have
Revised: Hence we have

Original: [Equation]
Revised: [Equation]