Need help with the inequality problem

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FriendlyAsian

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Let x, y be positive real numbers satisfying x2+ y3 [MATH]\geq [/MATH] x3+ y4. Prove that x3+ y3[MATH]\leq[/MATH] 2.
I don't know where to start with. Can you guys give me ideas?
 
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well as these are positive integers you should be suspicious that there can't be too many pairs that are going to satisfy this inequality,
as the exponents are larger on the side of the equation that is said to be smaller.

In fact it looks like there is only 1 pair of positive integers that satisfies this inequality.
Find that and you're done.
 
FriendlyAsian said:
it seems like you need to prove that both x and y are smaller than 1
Click to expand...
Are you sure that your statement is true? Can you prove that if x and y are positive but less than 1 (can x and/or y be 1?), then the results follow. Showing that is not too difficult. Give it a try.
 
It really is not that hard.
Here is a start.
If 0<x<1 then x^2> x^3 and x^3<1
Can you go from here.
Remember this is just showing that your theorem is true if x and y are less than 1, but positive.
 
but you should consider the fact that if x^2+ y^3>x^3+y^4 then it is not neccessary to have both x and y< 1
x can be bigger than 1 and y can be smaller than 1
That's where i'm stuck!
 
FriendlyAsian said:
but you should consider the fact that if x^2+ y^3>x^3+y^4 then it is not neccessary to have both x and y< 1
x can be bigger than 1 and y can be smaller than 1
That's where i'm stuck!
Click to expand...
Let's see how far you can get with x and y both less than 1. Then we will consider other cases.
 
Maybe try proving the contrapositive: If x^3+ y^3 > 2 then x^2+ y^3 < x^3+ y^4.
Case1: Assume x>1
Case2: Assume y>1

Also maybe factoring the original may help. I will look at this more closely later today.
 
I don't think proving the contrapositive would solve the problem since you cannot reverse the inequality.
I'm trying to apply the AM-GM inequality to the problem and it seems to be working!
 
FriendlyAsian said:
I don't think proving the contrapositive would solve the problem since you cannot reverse the inequality.
I'm trying to apply the AM-GM inequality to the problem and it seems to be working!
Click to expand...
And why can't you reverse the inequalities signs? Is the negation of a<b, not a>b??
 
Hmm...I'm not sure but you can simply understand that it only works if there's an "equal" in the inequality??? If not it's false in some case, isn't it?
 
Can you check for errors in my solution? I accidentally came up with this while eating :) . Thanks!
x2+y3 [MATH]\geq [/MATH] y3+x4
<=> x2+ y3+ y2 [MATH]\geq [/MATH]x3+y4+y2[MATH]\geq [/MATH]x3+2y3
<=>x2+y2[MATH]\geq [/MATH]x3+y3 (1)
Since x and y are positive real numbers, we have:
(x2+y2+4)/3[MATH]\geq [/MATH](x3+ y3+ 4)/3 [MATH]\geq [/MATH]x2+y2(AM-GM)
<=>4/3[MATH]\geq [/MATH]2(x2+y2)/3
<=>2[MATH]\geq [/MATH]x2+y2 (2)
From (1) and (2) =>2 [MATH]\geq [/MATH]x3+y3 (Q.E.D)
"=" happens only when x=y=1
 
It is not an error, but

[MATH]y^4 + y^2 \ge 2y^3[/MATH]
is unsupported. So you need this lemma.

[MATH]y^2 \ge 0 \le (y - 1)^2 \implies y^2(y - 1)^2 \ge 0 \implies y^2(y^2 - 2y + y) \ge 0 \implies[/MATH]
[MATH]y^4 - 2y^3 + y^2 \ge 0 \implies y^4 + y^2 \ge 2y^3.[/MATH]
I have the same problem in the AM-GM line. Obviously

[MATH]\dfrac{x^3 + y^3 + 4}{3} \ge xy\sqrt[3]{4}.[/MATH]
But it is not obvious to me that

[MATH]xy\sqrt[3]{4} \ge x^2 + y^2.[/MATH]
I am not saying that proposition is wrong. I am just wondering whether you need to demonstrate it.
 
Sorry. I didn't have much time to type it all out, but here's the fully demonstration:
We have: x^3+1+1[MATH]\geq[/MATH]3x and y^3+1+1[MATH]\geq[/MATH]3y
Adding the two inequalities, we get (x^3+y^3+4)/3 [MATH]\geq[/MATH]x+y
Hence, we need to prove x+y[MATH]\geq[/MATH]x^2+y^2
It's obvious as (x2+y2+x+y)/2 [MATH]\geq [/MATH] (x3+y3+x+y)/2 [MATH]\geq[/MATH]x2+y2
=>x+y[MATH]\geq[/MATH]x2+y2
 
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