Solve the inequality $\sqrt{4x+1}+\sqrt{x+1}<x+1$

Question

Can you show the steps for solving this inequality: $$\sqrt{4x+1}+\sqrt{x+1}<x+1$$

Condition: $x \geq -1/4$ and $x\geq -1$.

I'm stuck here:

$$2\cdot \sqrt{(4x+1)(x+1)}<x^2-3x-1$$

Answer 1

We have $$ x\geq -\frac{1}{4} \;, $$ and after squaring one times (all terms are non negative) we obtain $$ 4x+1+x+1+2\sqrt{x+1}\sqrt{4x+1}<x^2+2x+1 \;. $$ Simplifiying, we get $$ 2\sqrt{x+1}\sqrt{4x+1}<x^2-3x-1 \;. $$ For the right hand side to be nonnegative, we have $$ x\geq \frac{1}{2}(3+\sqrt{13}) \;. $$ Now we can square it again: $$ 4(x+1)(4x+1)<(x^2-3x-1)^2 \;. $$ Simplification gives $$ {x}^{4}-6\,{x}^{3}-9\,{x}^{2}-14\,x-3 >0 \;, $$ and this inequality gives the solution: $$ x>7.46 \ldots $$