Prove if $x\gt3$ then $1\ge\frac{3}{x(x-2)}$.

Question

I tried to prove it by contradiction.

Suppose it is not true that $1\ge\frac{3}{x(x-2)}$, so $1\lt\frac{3}{x(x-2)}$. Then $\frac{3}{x(x-2)}-1\gt0$. Multiply both sides of $\frac{3}{x(x-2)}-1\gt0$ by ${x(x-2)}$.

$(\frac{3}{x(x-2)}-1\gt0)({x(x-2)}\gt0(x(x-2)$

${3-(x(x-2)\gt0}$

${3-x^2-2x\gt0}$

${-x^2-2x+3\gt0}$

${-1(x^2+2x-3)\gt0}$

$-1\frac{(x-1)(x+3)}{-1}\gt0/-1$

${(x-1)(x+3)\lt0}$

At this point I really do not know what to do after this point or if I really even went about it the right way. Thank you for the help.

Answer 1

From $$x-3\geq 0$$ follows: $$x-1\geq 2$$ $$(x-1)^2\geq 4$$ $$x^2-2x+1\geq 4$$ $$x^2-2x\geq 3$$ $$x(x-2)\geq 3$$ so $$1\geq \frac{3}{x(x-2)}$$

Answer 2

$x > 3 \implies x-2 > 1 \implies x(x-2) > 3 \implies 1 > \dfrac{3}{x(x-2)} $

Answer 3

Note that if $x>3\implies x-2>1$ then $\dfrac1x<\dfrac13$ and $\dfrac1{x-2}<1$ so $$\frac3{x(x-2)}<\frac3{3(3-2)}=1$$

Answer 4

Why do you make things more complex than they are?

The function $x(x-2)$ is increasing and positive on $(2,+\infty)$, hence $\;\dfrac2{x(x-2)}\;$ is decreasing and positive on this interval. Furthermore $f(3)=1$…