Please help with this inequality (x^2-3x+5)^2>4(x-1)

[georgenik8]

I can't seem to figure out this inequality. I found it online when I was searching for exercises in algebra. The exercise asks to prove that for every x in R this inequality is valid: (x^2-3x+5)^2>4(x-1)
I know it's valid for x<= 1 because something squared is always positive or 0 and if x is <= 1 then the right side would be negative and also it cannot be 0 because x^2-3x+5 is always positive.
Thanks in advance.

 

[stapel]

I can't seem to figure out this inequality. I found it online when I was searching for exercises in algebra. The exercise asks me to:

Prove that, for every x in R, this inequality is valid:

. . . . .(x^2 - 3x + 5)^2 > 4(x - 1)

I know it's valid for x <= 1, because something squared is always positive or 0; and if x <= 1, then the right side would be negative. Also, it cannot be 0 because x^2 - 3x + 5 is always positive.

Can you determine the minimum value of the squared quadratic? What have you learned about the comparative "speeds" of growth of quadratics (and quartics) versus linear functions? Have you considered subtracting the right-hand side over to the left-hand side, and looking at the quartic that results?

Thank you!

 

[Steven G]

I can't seem to figure out this inequality. I found it online when I was searching for exercises in algebra. The exercise asks to prove that for every x in R this inequality is valid: (x^2-3x+5)^2>4(x-1)
I know it's valid for x<= 1 because something squared is always positive or 0 and if x is <= 1 then the right side would be negative and also it cannot be 0 because x^2-3x+5 is always positive.
Thanks in advance.

My first step would be, as already mentioned, in these type of problems is to get one side equal to 0 and see why the other side is always positive or always negative.

You have to try something. Show us your work and we can guide you from there.

 

[georgenik8]

It actually didn't occur to me to tackle it like that. Although i found the min value of the squared quadratic which is point K(3/2, 121/16) I have trouble understanding what how to handle the -4(x-1), but I will try to figure out a way to solve it.
Thank you.

For some reason i can't post picture of my work so I had to type just the results. Sorry.

 

[Steven G]

It actually didn't occur to me to tackle it like that. Although i found the min value of the squared quadratic which is point K(3/2, 121/16) I have trouble understanding what how to handle the -4(x-1), but I will try to figure out a way to solve it.
Thank you.

For some reason i can't post picture of my work so I had to type just the results. Sorry.

Finding the min for the left hand side will not help much and here is why. We know lhs >0 and say the min value is k, so if we can show that 4(x-1)<k for all x then we are done. But 4(x-1) can be as large as we want (it is a line whose slope is not zero). Just let x-1=k or x-1=k/2 and you'll see that 4(x-1)>k.

Just get one side to equal zero and factor, or find the min of one side while the other side is 0 (or some constant)