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?

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 :
$x-4x\frac{x}{\sqrt{x} +x} +3x(\frac{x}{\sqrt{x} +x})^2 - 3(\frac{x}{\sqrt{x} +x})^2 <0$
