solving inequalities with fractions on both sides Solve this inequality: $\frac{x^2 -2}{2} < \frac{6x^2 -8x - 1}{x+5}$.
My solution: 
Multiply $(x+5)^2$ on both sides: 
$\frac{(x^2 -2)(x+5)^2}{2} = \frac{x^4 + 10x^3 + 23x^2 -20x -50}{2}$
$(6x^2 -8x - 1)(x+5) = 6x^3 + 22x^2 -41x - 5 $
$\frac{x^4 + 10x^3 + 23x^2 -20x -50}{2} < 6x^3 + 22x^2 -41x - 5$
$0.5x^4 - x^3 - 10.5x^2 + 31x - 20 < 0 $
I don't think doing this way will lead to solving the question....

Solution ($\color{red}{Edited}$):
Given 
$$\frac{(x^2 -2)}{2} < \frac{(6x^2 -8x - 1)}{(x+5)}$$
1. Multiply both sides by 2$(x+5)^2$ ($\color{red}{Since\;}$$\color{red}{(x+5)^2\ge 0,\;the\;inequality\;is\;preserved}$):
$$(x+5)^2(x^2-2)<2(x+5)(6x^2-8x-1)$$
2. Expand and collect like terms:
$$x^4-2x^3-21x^2+62x-40<0$$
3. Factor the polynomial:
$$(x+5)(x-1)(x-2)(x-4)<0$$
4. Draw the critical points on the real line

We now can write the domain as the union 
$$D=D_1 \cup D_2 \cup D_3 \cup D_4 \cup D_5=(-\infty,-5) \cup (-5,1) \cup (1,2) \cup (2,4) \cup (4,\infty)$$
6. By studying the sign of each factor, answer the following: Is the left-hand side negative (less than zero)?

  
*
  
*On $D_1$: $$ (-)(-)(-)(-)=(+)<0$$ which is false.
  
*On $D_2$: $$ (-)(-)(-)(+)=(-)<0$$ which is true.
  
*On $D_3$: $$ (+)(-)(-)(+)=(+)<0$$ which is false.
  
*On $D_4$: $$ (+)(+)(-)(+)=(-)<0$$ which is true.
  
*On $D_5$: $$ (+)(+)(+)(+)=(+)<0$$ which is false.
  

Thus the solution is 
$$x=(-5,1)\cup (2,4)$$
as shown in the following plot

