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

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up vote 2 down vote accepted

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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I've changed $-13$ to $+13$ in two spaces (probably a typo). BTW the same can be done with less intermediate steps as $(x^2-6x+13)^2-(4x-12)^2=(x^2-10x+25)(x^2-2x+1)=(x-5)^2(x-1)^2$; which is obtained using the formula $a^2-b^2=(a-b)(a+b)$ for $a=x^2-6x+13$ and $b=4x-12$. – Martin Sleziak Jul 23 '12 at 13:28
@MartinSleziak thanks – draks ... Jul 23 '12 at 13:29

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.

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