Inequality

Problem from Facebook/Mathematics 2010-05-02

If a, b, c are non negative real numbers such that a+b+c =1, then show that a/(1+bc) +b/(1+ca) + c/(1+ab) >=9/10.

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Solution 1

(a/α + b/β + c/γ)(aα + bβ + cγ) =
= a² + β/α·ab + γ/α·ac + α/β·ba + b² + γ/β·bc +
+ α/γ·ca + β/γ·cb + c² =
= a² + b² + c² + (β/α + α/β)·ab + (γ/β + β/γ)·bc +
+ (γ/α + α/γ)·ca ≥ a² + b² + c² + 2ab + 2bc + 2ca =
= (a + b + c)², which proves your inequality.

Here we have used α/β + β/α = (β² + α²)/(αβ) =
= [(β – α)² + 2αβ] / (αβ) ≥ 2, cycl.

Next, using the inequality gm ≤ am, Σ a(1+bc) = a + b + c +  3abc = 1 + 3abc ≤ 1 + 3·[(a + b + c)/3]³ = 10/9.

Finally Σ [a/(1+bc)] ≥ (a + b + c)² / Σ a(1+bc) ≥ 9/10.

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Solution 2

Let f(a,b,c) = a/(1+bc) + b/(1+ca) + c/(1+ab) =
= a²/(a+abc) +b²/(b+abc) + c²/(c+abc).
By the conditions a + b + c = 1 and a, b, c ≥ 0 the allowed region Δ for (a,b,c) in ℝ³ is a tetrahedron with corners (0,0,0), (1,0,0), (0,1,0), and (0,0,1…).

Now, on the face Σ of Δ with corners (1,0,0), (0,1,0), and (0,0,1), i e for a + b + c = 1, and and a, b, c ≥ 0, we have 0 ≤ abc ≤ 1/27, with largest value for a = b = c = 1/3.
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Proof for this: By the condition a + b + c = 1,
abc = ab(1 – a – b).
Let g(ab) = abc = ab(1 – a – b) = ab – a²b – ab² in the face Σ₁ with corners (0,0,0), (1,0,0), (0,1,0), or a ≥ 0, b ≥ 0, a+b ≤ 1.

∂g/∂a = b – 2ab – b² = b(1 – 2a – b)
∂g/∂b = a – a² – 2ab = a(1 – a – 2b)
From this, inside the region Σ₁, ∂g/∂a = ∂g/∂b = 0 iff a = b = 1/3. At this point g(a,b) = 1/27.
Futher, on the hypothenuse of Σ₁, a + b = 1, so g(a,b) =
= a(1 – a)·(1 – (a+b)) = 0. On the other sides of Σ₁, g(a,b) = 0, since a = 0 or b = 0. Since smallest/largest values occur on the boundary of Σ₁ or at points inside Σ₁ where ∂g/∂a = ∂g/∂b = 0, 0 ≤ abc ≤ 1/27. qed
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By this little theorem, in particular abc ≤ 1/27, so
f(a,b,c) = a²/(a+abc) +b²/(a+abc) + c²/(c+abc) ≥
≥ a²/(a+1/27) +b²/(a+1/27) + c²/(c+1/27) =
= 27( a²/(27a+1) +b²/(27b+1) + c²/(27c+1) ).
For simplicity, set A = 27a, B = 27b, and C = 27c, and let
h(A,B,C) = A²/(A+1) +B²/(B+1) + C²/(C+1), so that f(a,b,c) = h(A,B) / 27.

On Σ, A + B + C = 27, i e C = 27 – A – B, so (using the chain rule and ∂C/∂A = ∂C/∂B = -1 ),
∂h/∂A = A(A + 2)/(A+1)² – C(C + 2)/(C+1)², and
∂h/∂B = B(B + 2)/(B+1)² – C(C + 2)/(C+1)²
From this ∂h/∂A = ∂h/∂B = 0 inside Σ₁ iff A = B = C = 9.
On the hypothenuse of Σ₁, A + B = 27, and C = 0 so
h(A,B,C) = A²/(A+1) +(27 – A)²/(28 – A), which takes the smallest value (i e for 0 ≤ A ≤ 27) 729/29 when A = 27/2, ie at the middle of the side.
On one of the other sides of Σ₁, B = 0,
h(A,B,C) = A²/(A+1) + (27-A)²/(28-A), which takes the smallest value 729/29 for A = 27/2, i e at the middle of side.
Similarly and same result for the other side of Σ₁, A = 0.

So, the smallest value of h(A,B,C) on the boundary of Σ₁ is 729/27, and the extreme value inside Σ₁ is h(9,9,9) = 243/10 < 729/29, so 243/10 is the smallest value, and so h(A,B,C) ≥ 243/10 for A + B + C = 27, and A, B, C ≥ 0.

Finally f(a,b,c) ≥ h(A,B) / 27 ≥ 9/10 for a + b + c = 1, and a, b, c ≥ 0. QED

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