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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?

minthao_2011
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2 Answers2

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a picture

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.

eccstartup
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  • Of course, you must make sure that $t \ge 0$. – eccstartup Jul 11 '13 at 04:35
  • Viva China ;-) +1 – Mikasa Aug 20 '13 at 06:39
  • @eccstartup Good answer! but note that more precisely we should consider $t>0$ Furthermore for $t=0$ we have $P(0)=4$ hence the leftmost must be above the axis. If you draw the graph for $t\in\mathbb{R}$ the leftmost must be above the axis too because the graph behave like $2t^8$ as $t\to\pm\infty$. – Etemon Jun 16 '21 at 21:07
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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

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

Calvin Lin
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