Prove the inequality $x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$. I want to prove $$x-4x \frac{x}{\sqrt{x}+x} +3x  \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$$
for $x>0$ real number.
I tried supposing that  $$x-4x \frac{x}{\sqrt{x}+x} +3x  \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 \geq 0$$ holds to get a contradiction but I couldn't find the solution.
Can you help me proceed?
 A: EDIT: The question was updated, so here's my new answer.
Notice the LHS is

and since $x>0$, it's clear that it is $<0$ for all $x$.

Old answer: This is false. Indeed, since $x>0$, we have
\begin{align*}
   &x-4x \frac{1}{\sqrt{x}+x} +3x \left( \frac{1}{\sqrt{x}+x} \right)^2 -3 \left(\frac{1}{\sqrt{x}+x} \right)^2 >0\\
  \iff & (x+\sqrt x)^2\left(x-4x \frac{1}{\sqrt{x}+x} +3x \left( \frac{1}{\sqrt{x}+x} \right)^2 -3 \left(\frac{1}{\sqrt{x}+x} \right)^2 \right) >0\\
  \iff & x^3 + 2x^{5/2}-3x^2-4x^{3/2}+3x-3 >0\\
 \iff & (\sqrt x+1)(x^{5/2}+x^2-4x^{3/2}+3x^{1/2}-3)>0,
\end{align*}
but this isn't true for all $x$, e.g. put $x=1$ and the quintic becomes $-2<0$.
A: Hint:  Simplify that LHS to  $\displaystyle -\frac{2x}{1+\sqrt x}< 0$, which is obvious for $x>0$.
A: Answer :
$x-4x\frac{x}{\sqrt{x} +x} +3x(\frac{x}{\sqrt{x} +x})^2 - 3(\frac{x}{\sqrt{x} +x})^2 =1-4\frac{x}{\sqrt{x} +x} +3(\frac{x}{\sqrt{x} +x})^2 - 3\frac{x}{(\sqrt{x} +x)^2 }=  \frac{(\sqrt{x} +x)^2 - 4x(\sqrt{x} +x) +3x^2 - 3x}{(\sqrt{x} +x)^2} =\frac{x+2x\sqrt{x} +x^2 - 4x\sqrt{x} - 4x^2 +3x^2 - 3x}{(\sqrt{x} +x)^2} =\frac{-2x-2x\sqrt{x}}{(\sqrt{x} +x)^2}=\frac{-2\sqrt{x}}{\sqrt{x} +x} <0$
Finally :
$x-4x\frac{x}{\sqrt{x} +x} +3x(\frac{x}{\sqrt{x} +x})^2 - 3(\frac{x}{\sqrt{x} +x})^2 <0$
