Algebra inequality.

Hi all,

no I am not asking a question, I am just answering for someone off this site.   

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$\displaystyle \frac{(1-x)^2}{\sqrt{x}}\ge 4\sqrt{x}(x-1)\\$

You cannot divide by 0 and you cannot take the square root of a negative number so  X>0

sqrt x is positive so when I multiply both sides by sqrtx the sign will stay the same.

$\displaystyle \frac{(1-x)^2}{\sqrt{x}}*\sqrt{x} \ge 4\sqrt{x}\sqrt{x}(x-1)\\ (1-x)^2\quad \ge 4x(x-1)\\ x^2-2x+1 \ge 4x^2-4x\\ -3 x^2+2x+1 \ge 0\\ 3x^2-2x-1\le0\\ \qquad consider\;\; \\ \qquad3x^2-2x-1=0\\ \qquad x=\frac{2\pm\sqrt{4+12}}{6}=\frac{2\pm4}{6}=\frac{1\pm2}{3}=\frac{-1}{3}\;\;or\;\;1\\ \qquad so\;\;3x^2-2x-1=(x+\frac{1}{3})(x-1)\\ (x+\frac{1}{3})(x-1)\le 0$

This is a concave up parabola so the bit between the roots will lie under the x axis. this means that its value, between the roots, will be negative.  Sketch  y=x^2-2x-3 so try and really appreciate what I am talking about.

so  

$\frac{-1}{3}\le x\le1$

BUT we already know that x must be bigger than 0 so

\(0

Edit:   This lat bit of LaTex refusing to display properly so I will display it as a picure

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Here is a representation.  

You want where the red graph,  is under the green     LHS <= RHS    of the original inequality.

LaTes:

\displaystyle \frac{(1-x)^2}{\sqrt{x}}\ge 4\sqrt{x}(x-1)\\

\displaystyle \frac{(1-x)^2}{\sqrt{x}}*\sqrt{x} \ge 4\sqrt{x}\sqrt{x}(x-1)\\
 (1-x)^2\quad  \ge 4x(x-1)\\
 x^2-2x+1 \ge 4x^2-4x\\
-3 x^2+2x+1 \ge 0\\
3x^2-2x-1\le0\\
\qquad consider\;\; \\
\qquad3x^2-2x-1=0\\
\qquad x=\frac{2\pm\sqrt{4+12}}{6}=\frac{2\pm4}{6}=\frac{1\pm2}{3}=\frac{-1}{3}\;\;or\;\;1\\
\qquad so\;\;3x^2-2x-1=(x+\frac{1}{3})(x-1)\\

(x+\frac{1}{3})(x-1)\le 0