# How to solve the irrational inequality?

Solve the inequality $$\dfrac{2x^4+2x^2}{\sqrt{x+1}}+(x+2)\sqrt{x+1}>x ^3 + 2x^2 + 5x.$$ I tried. By putting $t = \sqrt{x+1}$, we have $$2t^8-t^7-8t^6+t^5+15t^4-4t^3-11t^2+4t+4>0.$$ Using Maple, I got $$(2t^2-t-2)(t^2-t-1)(t^4+t^3-t^2-t+2)>0.$$ How to solve the given inequality with another way?

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I'm not sure if you're asking for a completely different method, or don't mind continuing what you have. – Calvin Lin Jul 11 '13 at 4:27

## 2 Answers

Try to make sure the coefficient of the largest degree of $t$ is positive.

Then, solve all the roots of the polynomial $P(t)=0$.

Then draw a picture as above.

Draw a axis.

Then draw a thread goes through the line from the roots. If there is multiple root, goes through the line and goes back. Right to left. Rightmost must be above the axis.

The positive intervals are above the axis.

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Of course, you must make sure that $t \ge 0$. – eccstartup Jul 11 '13 at 4:35
Viva China ;-) +1 – Babak S. Aug 20 '13 at 6:39

Hint: You also want the condition that $t \geq 0$.

Hint: The quartic polynomial $t^4 + t^3 - t^2 - t + 2 = (t^2 + 0.5 t -1)^2 + 0.75t^2 + 1$ is always positive for all real $t$.

You can continue from what you have (assuming your calculations are correct).

I'm not sure if this helps, but your inequality can be simplified to

$$(x^2-x+2)(2x^2+2x+1) > x(x^2+2x+5) \sqrt{x+1}$$

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