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Math Help - Find the Real solutions

  1. #1
    Newbie Sorombo's Avatar
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    Find the Real solutions

    Find the real solutions for:

    \sqrt x+\sqrt y +\sqrt {z-2} +\sqrt u+ \sqrt v = x+y+z+u+v
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  2. #2
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    Awetuouncsygg
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    \sqrt x+\sqrt y +\sqrt {z-2} +\sqrt u+ \sqrt v = x+y+z+u+v Has no real solutions.

    1.
    x-\sqrt x+y-\sqrt y+(z-2) -\sqrt {z-2}+u -\sqrt u+v- \sqrt v+2 =0 \Rightarrow (x-\sqrt x+\frac{1}{4})+(y-\sqrt y+\frac{1}{4})+[(z-2)-\sqrt {z-2}+\frac{1}{4}]+(u-\sqrt u+\frac{1}{4})+(v-\sqrt v+\frac{1}{4})+2-\frac{5}{4}=0 \Rightarrow (\sqrt x-\frac{1}{2})^2+(\sqrt y-\frac{1}{2})^2+(\sqrt {z-2}-\frac{1}{2})^2+(\sqrt u-\frac{1}{2})^2+(\sqrt v-\frac{1}{2})^2+\frac{3}{4}=0
    Impossible because:
    (\sqrt x-\frac{1}{2})^2 \geq 0, (\sqrt y-\frac{1}{2})^2 \geq 0 , (\sqrt {z-2}-\frac{1}{2})^2 \geq 0 , (\sqrt u-\frac{1}{2})^2 \geq 0,  (\sqrt v-\frac{1}{2})^2 \geq 0 and \frac{3}{4} > 0.

    2.
    \frac{\sqrt x+\sqrt y +\sqrt {z-2} +\sqrt u+ \sqrt v}{5}\leq \sqrt{\frac{x+y+z-2+u+v}{5}}\Rightarrow
     \frac{x+y+z+u+v}{5}\leq \sqrt{\frac{x+y+z-2+u+v}{5}}\Rightarrow  \frac{S^2}{25} \leq \frac{S-2}{5}\Rightarrow S^2-5S+10\leq 0\Rightarrow  \left ( S-\frac{5}{2} \right )^2+\frac{15}{4}\leq 0
    Impossible because:
     \left ( S-\frac{5}{2} \right )^2\geq 0 and \frac{15}{4}>0.

    S=x+y+z+u+v
    I used the inequality between the root mean square and the arithmetic mean of numbers x, y, z, u and v.
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