Well, (x-3)^2 is definitely positive and +4 that's no doubt positive as well, doesn't seem to matter me :P
I kinda get it, but I guess it's necessary for me to explain so much text in the working as well?
Sorry... I still don't see how it applies to the problem. So when x < 3 it's negative and x > 3 it's positive, then? How does this relate to the problem at all?
Okay, I factorized and I got (x-5)^2(x-3) \geq 0, since (x-5)^2 is always positive, so I guess that leaves me to x > 3. What does this means basically?
It's \frac{x^2 - 6x - 13}{2(x-3)} right? Hopefully I did not calculate wrongly. If I solve x^2-6x-13=0, I'll get 3 - \sqrt{22} and 3 + \sqrt{22}. Putting them into x makes it zero, that's not what I want to prove?
I got this y - 5 = \frac{(x-3)^2 - 11}{2(x-3)}, which doesn't give any minimum value when x = 3, but by splitting it up I get \frac{x-3}{2} - \frac{11}{2(x-3)}, is that what I'm supposed to do? I still don't get the 2 values you got...
Homework Statement
If x is real and y=\frac{x^2+4x-17}{2(x-3)}, show that |y-5| \geq 2Homework Equations
The Attempt at a Solution
Sorry... Absolutely no idea. I tried to substitute y into the left side to prove that -2 \leq y - 5 \leq 2 but I can't. Anything I should know to do this?
I think I got it, by using \frac{a+b}{2}>\sqrt{ab}, etc and by multiplying a and b respectively then adding up all together made it. Thanks for all the tips!
Hmm, thanks for your tip off, but I still can't seem to make it...
The 2 is always appearing on the right side of the inequality, but as of the question it's on the right side, if I expand (a-b)^2 the 2 sticks together with ab instead of a^3...
Homework Statement
If a, b and c are distinct positive numbers, show that
2 (a^3 + b^3 + c^3) > a^2b + a^2c + b^2c + b^2a + c^2a + c^2b
Homework Equations
The Attempt at a Solution
I have tried to expand from (a+b+c)^3 > 0, also tried (a+b)^3 + (b+c)^3 + (c+a)^3 > 0, and then...