Solving Exponential inequalities $ ( x^2 + x +1) ^{ \frac{x+5 }{x+2}} \geq ( x^2 + x +1) ^3 $

Question

I have an inequality: $( x -3)^{2x^2 - 7x } \geq 1 $. Somewhere I found a solution in which we consider two cases: $ x -3 > 1$ and $ 0 < x-3 <1 $. My question is: can we consider the case when $ x -3 = 1$?

Can we do it also in inequality: $ ( x^2 + x +1) ^{ \frac{x+5 }{x+2}} \geq ( x^2 + x +1) ^3 $ and what are the solutions ?

Answer 1

Yes, you must consider the case $x-3=1$ because it can give an additional solution and indeed, it gives the solution $4$.

The second problem.

1. $x^2+x+1=1$ gives $x=0$ or $x=-1$, which valid with the domain of $\frac{x+5}{x+2}$;

2. $x^2+x+1>1$ and $\frac{x+5}{x+2}\geq3$;

$x<-1$ or $x>0$ and $\frac{x+5-3x-6}{x+2}\geq0$

$x<-1$ or $x>0$ and $\frac{2x+1}{x+2}\leq0$

$x<-1$ or $x>0$ and $-2<x\leq-\frac{1}{2}$

$-2<x<-1.$

3. $0<x^2+x+1<1$ and $\frac{x+5}{x+2}\leq3$

$-1<x<0$ and $x\in(-\infty,-2)\cup\left[\frac{1}{2},+\infty\right)$, which is nothing.

Finally we obtain: $$(-2,-1]\cup\{0\}$$

Answer 2

In your question, the two cases should be

$ x -3 \ge 1$ and $ 0 < x-3 <1 $,instead of

$ x -3 > 1$ and $ 0 < x-3 <1 $

as equality is included in your question. So there is no need of separate condition

And yes, conditions would be same for other question as well