If a,b,c>0 and a+b+c=3 prove that a/(1+b²)+b/(1+c²)+c/(1+a²) ≥ 3/2?

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  • atsuo
    Lv 6
    5 months ago
    Favorite Answer

    I think simpler proof may exist .

    Condition : a,b and c > 0 and a + b + c = 3

    a/(1 + b^2)

    = a(1 + b^2)/(1 + b^2) - ab^2/(1 + b^2)

    = a - ab^2/(1 + b^2)

    We know 1 + b^2 - 2b = (1 - b)^2 ≧ 0 , so

    1 + b^2 ≧ 2b

    1/(1 + b^2) ≦ 1/(2b)

    -1/(1 + b^2) ≧ -1/(2b)

    -ab^2/(1 + b^2) ≧ -ab^2/(2b) = -ab/2

    So

    a/(1 + b^2)

    = a - ab^2/(1 + b^2) ≧ a - ab/2

    Similarly ,

    b/(1 + c^2) ≧ b - bc/2 and c/(1 + a^2) ≧ c - ac/2 .

    Add above three inequalities , we can find

    a/(1 + b^2) + b/(1 + c^2) + c/(1 + a^2) ≧ a + b + c - (1/2)(ab + bc + ca)

    a/(1 + b^2) + b/(1 + c^2) + c/(1 + a^2) ≧ 3 - (1/2)(ab + bc + ca) ---(#1)

    Next , we know

    a^2 + b^2 + c^2 - (ab + bc + ac) = (1/2)(a-b)^2 + (1/2)(b-c)^2 + (1/2)(a-c)^2 ≧ 0

    So

    ab + bc + ac ≦ a^2 + b^2 + c^2

    3(ab + bc + ac) ≦ a^2 + b^2 + c^2 + 2(a^2 + b^2 + c^2) = (a + b + c)^2 = 9

    ab + bc + ac ≦ 3

    -(1/2)(ab + bc + ac) ≧ -3/2 ---(#2)

    By (#1) and (#2) ,

    a/(1 + b^2) + b/(1 + c^2) + c/(1 + a^2) ≧ 3 - 3/2 = 3/2

    So given inequality is proved .

  • 5 months ago

    fao Captain Matticus, LandPiratesInc

    The feasible region in an a-b-c coordinate system is an equilateral triangle. Your work shows that the

    inequality is true at the vertices, the mid-points of the sides and the centroid. The LHS of the inequality is a non-linear function of a,b,c and so it does not follow that the inequality holds at interior points. This is an olympiad standard problem so expect it to require some ingenuity.

  • All we have to do is check our extreme cases

    a = 0 , b = 0 , c = 3

    a = 0 , b = 3 , c = 0

    a = 3 , b = 0 , c = 0

    a = 0 , b = 1.5 , c = 1.5

    a = 1.5 , b = 0 , c = 1.5

    a = 1.5 , b = 1.5 , c = 0

    a = 1 , b = 1 , c = 1

    0/(1 + 0) + 0/(1 + 9) + 3/(1 + 0) = 3

    0/(1 + 9) + 3/(1 + 0) + 0/(1 + 0) = 3

    3/(1 + 0) + 0/(1 + 0) + 0/(1 + 0) = 3

    0/(1 + 2.25) + 1.5/(1 + 2.25) + 1.5/(1 + 0) = 1.5/3.25 + 1.5 = 6/13 + 3/2 = 12/26 + 39/26 = 51/26

    1.5/(1 + 0) + 0/(1 + 2.25) + 1.5/(1 + 2.25) = 51/26

    1.5/(1 + 2.25) + 1.5/(1 + 0) + 0/(1 + 2.25) = 51/26

    1/(1 + 1) + 1/(1 + 1) + 1/(1 + 1) = 1/2 + 1/2 + 1/2 = 3/2

    If you can think of some other extreme cases, I'm all ears.

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