How do i prove this inequality related to quadratic equations?

Question

For any real positive value of $x$, show that $3-x\not>\frac{7}{x+2}$.(Using properties of quadratic equation)

I have not posted any attempt because i have no idea where to start from. I have spent too much time trying to figure out where to start from and how to proceed. I would be grateful for any help i receive.

Answer 1

If possible, $3-x > \frac7{x+2}$. Then $$(3-x)(x+2)>7\\3x+6-x^2-2x>7\\x^2-x+1<0\\x(x-1)<-1<0$$

Therefore we must have $0<x<1$. Therefore, $x^2>0$ and $1-x >0$ which implies $x^2+1-x>0$ which gives a contradiction.

Answer 2

Let $$f (x)=(3-x)-\frac {7}{x+2}$$

$$=\frac {-x^2+x-1}{x+2} .$$

the discriminant of numerator is $$\Delta=1-4 <0$$ thus

$$(\forall x\in\mathbb R) \;\; -x^2+x-1 <0$$ and $$(\forall x>0)\; \;\; x+2>0. $$

You can finish.

Answer 3

For which values $3-x>\frac{7}{x+2}$?

If $3-x>\frac{7}{x+2}$ then $3-x-\frac{7}{x+2}>0 \rightarrow \frac{-x^2+x-1}{x+2}>0$

But $-x^2+x-1<0 \forall x \in \mathbb{R}$

So $\frac{-x^2+x-1}{x+2}>0$ only if $x+2<0 \rightarrow x<-2$

Answer 4

Hint:  given that $x \gt 0$ then $3-x\not\gt\frac{7}{x+2} \iff 3-x\le\frac{7}{x+2} \iff (3-x)(x+2)\le 7\,$.

• For $x \ge 3$ the LHS is negative, so the inequality is always satisfied.

• For $0 \lt x \lt 3$ it follows by AM-GM that $\require{cancel}\,(3-x)(x+2)\le\left(\frac{3-\bcancel{x}+\bcancel{x}+2}{2}\right)^2=\frac{25}{4}\,$.