Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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.

share|cite|improve this answer
    
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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.