Inequality problems I have Maths test tomorrow and was just doing my revision when I came across these two questions. Would anyone please give me a nudge in the right direction?
$1)$ If $x$ is real and $$y=\frac{x^2+4x-17}{2(x-3)},$$ show that $|y-5|\geq2$ 
$2)$ If $a>0$, $b>0$, prove that $$\left(a+\frac1b\right)\left(2b+\frac1{2a}\right)\ge\frac92$$
 A: Expression in (2) is 
$$
2ab+\frac{1}{2ab}+\frac{5}{2}
$$
Applying $AM \geq GM$ on the first two terms gives
$$
2ab+\frac{1}{2ab} \geq 2
$$
Substituting in the previous expression yields the given inequality.
A: Only the first problem.
By long division you have
$$y=\frac{x+7}2+\frac2{x-3}$$
and thus
$$y-5=\frac{x-3}2+\frac2{x-3}.$$
So if you substitute $t=\frac{x-3}2$, it suffices to show
$$\left|t+\frac1t\right|\ge2$$
for $t\ne 0$, which is the same as
$$t+\frac1t\ge2$$
for $t>0$.
There are many methods how to show this last inequality.
A: For the first problem:
Write it as 
$$
\begin{eqnarray}
\left(y-5\right)^2-4&=&\left(\frac{x^2+4x-17}{2(x-3)}-5\right)^2-4\\
&=&\left(\frac{x^2+4x-17-10x+30}{2(x-3)}\right)^2-4\\
&=&\left(\frac{x^2-6x+13}{2(x-3)}\right)^2-4\\
&=&\frac{(x^2-6x+13)^2 - 16(x-3)^2}{4(x-3)^2}\\
&=&\frac{169-156 x+62 x^2-12 x^3+x^4 - 16x^2+96x-144}{4(x-3)^2}\\
&=&\frac{x^4 -12x^3+46x^2 -60x+25}{4(x-3)^2}\\
&=&\frac{(x^2 -6x+5)^2}{4(x-3)^2}\\
&=&\frac{(x-5)^2(x-1)^2}{4(x-3)^2}\ge 0
\end{eqnarray}
$$
So only squares show up, hence it's positive.
