Find the Real solutions

• April 3rd 2011, 06:31 PM
Sorombo
Find the Real solutions
Find the real solutions for:

$\sqrt x+\sqrt y +\sqrt {z-2} +\sqrt u+ \sqrt v = x+y+z+u+v$
• April 3rd 2011, 09:10 PM
veileen
$\sqrt x+\sqrt y +\sqrt {z-2} +\sqrt u+ \sqrt v = x+y+z+u+v$ 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.