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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 $\left(a+\frac1b\right)\left(2b+\frac1{2a}\right)\ge\frac92$

3 Answers 3

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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.

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    @Marti$n$Sleziak tha$n$ks2012-07-23
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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.

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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.