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Math Help - pozitive real numbers

  1. #1
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    pozitive real numbers

    Let  x, y, z be pozitive real numbers satisfying x+y+z =1



     6xyz \leq \frac{x(y + z)}{4 - 9yz}+\frac{y(z + x)}{4 - 9zx}+\frac{z(x + y)}{4 - 9xy}
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    By AM–GM, we have


    3\sqrt{xy}(2-3\sqrt{xy})\ \le\ \frac{3\sqrt{xy}+(2-3\sqrt{xy})}2\ =\ 1

    and

    2+3\sqrt{xy}\ \le\ 2+\frac{3(x+y)}2\ =\ \frac{4+3(x+y)}2


    Also, note that 2-3\sqrt{xy} is positive (because, by AM–GM, \sqrt{xy}\le\frac{x+y}2<\frac12\,). So, multiplying the above two inequalities gives


    3\sqrt{xy}(4-9xy)\ \le\ \frac{4+3(x+y)}2


    Hence \frac{x+y}{3xy(4-9xy)}\ \ge \frac{2\sqrt{xy}}{3xy(4-9xy)}\ =\ \frac2{3\sqrt{xy}(4-9xy)}\ \ge\ \frac4{4+3(x+y)}

    Similarly, \frac{y+z}{3yz(4-9yz)}\ \ge \frac4{4+3(y+z)} and \frac{z+x}{3zx(4-9zx)}\ \ge \frac4{4+3(z+x)}.


    \therefore\ \frac{x+y}{3xy(4-9xy)}+\frac{y+z}{3yz(4-9yz)}+\frac{z+x}{3zx(4-9zx)}

    \ge\ \frac4{4+3(x+y)}+\frac4{4+3(y+z)}+\frac4{4+3(z+x)}

    \ge\ \frac9{\frac{4+3(x+y)}4+\frac{4+3(y+z)}4+\frac{4+3  (z+x)}4} (by inequality of arithmetic and harmonic means)

    =\ \frac{36}{12+6(x+y+z)}

    =\ 2

    Multiplying through by 3xyz will give the required inequality.
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