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Math Help - Inequality

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    Inequality

    Prove that if ab+bc+ca=1 than \frac{1-a^2}{1+a^2}+\frac{1-b^2}{1+b^2}+\frac{1-c^2}{1+c^2}\le\frac{3}{2}. ( a,b,c\in\mathbb{R})
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  2. #2
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    Quote Originally Posted by james_bond View Post
    Prove that if ab+bc+ca=1 than \frac{1-a^2}{1+a^2}+\frac{1-b^2}{1+b^2}+\frac{1-c^2}{1+c^2}\le\frac{3}{2}. ( a,b,c\in\mathbb{R})
    Who thinks these horrible things up (and why do I spend hours brooding over them)?

    I'll assume that a, b and c are all positive (the problem's hard enough as it is, and I don't want to be bothered with negative quantities).

    Notice first that \frac{1-a^2}{1+a^2} = \frac{2-(1+a^2)}{1+a^2} = \frac2{1+a^2}-1 (and similarly for b and c), so the inequality can be written

    . . . . . \frac1{1+a^2}+\frac1{1+b^2}+\frac1{1+c^2} \leqslant\frac94 . . .(1)

    If bc+ca+ab=1 then

    . . . . . 1+a^2 = (bc+ca+ab)+a^2 = (a+b)(a+c) . . .(2)

    (with similar inequalities for b and c).

    Also, a+b+c = (a+b+c)(bc+ca+ab) = (b+c)(c+a)(a+b) + abc . . .(3)

    (just multiply out both sides to verify that).

    By the arithmetic-geometric mean inequality,

    . . . . . abc = \sqrt{bc}\sqrt{ca}\sqrt{ab} \leqslant \frac{b+c}2\frac{c+a}2\frac{a+b}2 = {\textstyle\frac18}(b+c)(c+a)(a+b) . . .(4)

    (that's the only place where I assume that a, b and c are positive).

    It follows that

    . . . . . \frac1{1+a^2}+\frac1{1+b^2}+\frac1{1+c^2} = \frac1{(a+b)(a+c)} + \frac1{(b+c)(b+a)} + \frac1{(c+a)(c+b)} . . .by (2)

    . . . . . . . . . . . . . . . . . . . . . . .  = \frac{2(a+b+c)} {(b+c)(c+a)(a+b)}

    . . . . . . . . . . . . . . . . . . . . . . .  = \frac{2(b+c)(c+a)(a+b) + 2abc}{(b+c)(c+a)(a+b)} . . .by (3)

    . . . . . . . . . . . . . . . . . . . . . . .  = 2 + \frac{2abc}{(b+c)(c+a)(a+b)}

    . . . . . . . . . . . . . . . . . . . . . . .  \leqslant 2 + \textstyle\frac14 = \frac94 . . .by (4).

    By (1), that was what we wanted to prove.
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