Proving Inequality between Positive Real Numbers a and b: a(a+b) ≤ 2

[anemone]

Let a and b be positive real numbers such that $a^5+b^3 \le a^2+b^2$.

Prove that $a(a+b) \le 2$.

 

[anuttarasammyak]

Spoiler

Plot of the two equations for investigation
https://www.wolframalpha.com/input?i2d=true&i=plot+Power[x,5]+Power[y,3]-Power[x,2]-Power[y,2]=0++and+x\(40)x+y\(41)=2+for+0<x<2+0<y<2&lang=en
The crossing point is (1,1). Looking at first and second derivatives at this point and investigating the extremes would prove the statement though it may be beyond high school.

In more preliminary thoughts
$$x^5+y^3-x^2-y^2=0...(1)$$
has points (0,1) (1,0)
$$x(x+y)=2...(2)$$
has points (0,+\infty) (\sqrt{2},0)
They have only one common point (1,1)

$$\infty>1,\sqrt{2}>1$$
mean (1) and (2) touch at (1,1) but no crossing. So looking at xy plane graphs of (1) and (2), we know the inequality is satisfied.

 

[bob012345]

This was an interesting problem I enjoyed working on. Here is my solution.

Spoiler

We are given $$a^5+b^3 \le a^2+b^2$$ and wish to show if that then $$a(a+b) \le 2$$

add $a^2 + ab$ to both sides

$$a^2 + ab + a^5+b^3 \le a^2 + ab + a^2+b^2$$ then move some terms over

$$a(a + b) \le 2a^2 + ab +b^2 - a^5 - b^3$$ the expression on the RHS is some function

$$f(a,b) = 2a^2 + ab +b^2 - a^5 - b^3$$ which we can find the maximum of. Taking partials we get

$$\frac{\partial f(a,b)}{\partial a} = 4a + b -5a^4 = 0$$ and
$$\frac{\partial f(a,b)}{\partial b} = a +2b -3b^2 = 0$$we get

$$ a = 3b^2 -2b$$ and $$b = 5a^4 -4a$$ substituting we get this beast
$$75a^7 -120a^4 -10a^3 + 48a +7 = 0$$ which has real positive solutions $a=1$ and $a≈0.90173$ the latter gives $b≈-0.30812$ which is not allowed so we have $a=1$ and therefore $b=1$

this gives $$f(1,1) = 2a^2 + ab +b^2 - a^5 - b^3 = 2 + 1 + 1 - 1 -1 = 2$$ thus

$$a(a+b) \le 2$$ given the original constraint.

Mathcad 3D

https://www.math3d.org/TRHrXnwZS
[\SPOILER]

 

[anuttarasammyak]

Spoiler

Let us observe two (x,y) plots
$$x^5-x^2+y^3-y^2=0...(1)$$
$$x(x+y)=2...(2)$$
for x,y>0. From (2)
$$y=\frac{2}{x}-x...(3)$$
By substituting y in (1) by (3), we get x equation of crossing of (1) and (2). Multiplyng x^3, it is
$$(x-1)^2(x^6+2x^5+2x^4+4x^2+12x+8)=0...(4)$$
As x^6+2x^5+2x^4+4x^2+12x+8 >0 for x>0, we know at (x,y)=(1,1) (1) and (2) touch but they have no crossing points for x>0.
Thus (x,y) x>0 y>0 satisfying
$$x^5-x^2+y^3-y^2 \leq 0$$
also satisfy
$$x(x+y) \leq 2$$
Fig. https://www.wolframalpha.com/input?i2d=true&i=plot+Power[x,5]+Power[y,3]-Power[x,2]-Power[y,2]=0++and+x\(40)x+y\(41)=2+for+0<x<2+0<y<2&lang=en
QED

 

[Office_Shredder]

This was an interesting problem I enjoyed working on. Here is my solution.

Spoiler

We are given $$a^5+b^3 \le a^2+b^2$$ and wish to show if that then $$a(a+b) \le 2$$

add $a^2 + ab$ to both sides

$$a^2 + ab + a^5+b^3 \le a^2 + ab + a^2+b^2$$ then move some terms over

$$a(a + b) \le 2a^2 + ab +b^2 - a^5 - b^3$$ the expression on the RHS is some function

$$f(a,b) = 2a^2 + ab +b^2 - a^5 - b^3$$ which we can find the maximum of. Taking partials we get

$$\frac{\partial f(a,b)}{\partial a} = 4a + b -5a^4 = 0$$ and
$$\frac{\partial f(a,b)}{\partial b} = a +2b -3b^2 = 0$$we get

$$ a = 3b^2 -2b$$ and $$b = 5a^4 -4a$$ substituting we get this beast
$$75a^7 -120a^4 -10a^3 + 48a +7 = 0$$ which has real positive solutions $a=1$ and $a≈0.90173$ the latter gives $b≈-0.30812$ which is not allowed so we have $a=1$ and therefore $b=1$

this gives $$f(1,1) = 2a^2 + ab +b^2 - a^5 - b^3 = 2 + 1 + 1 - 1 -1 = 2$$ thus

$$a(a+b) \le 2$$

Mathcad 3D
View attachment 314293https://www.math3d.org/TRHrXnwZS
[\SPOILER]

Well hold on now. If you're going to find the maximum subject to the constraints, then you have to be concerned about the possibility that the maximum lies on the constraint, and the gradient is not zero there.

 

[bob012345]

Well hold on now. If you're going to find the maximum subject to the constraints, then you have to be concerned about the possibility that the maximum lies on the constraint, and the gradient is not zero there.

I'm not finding the maximum of the LHS subject to the original constraint but the maximum of the entire function on the RHS and that is less than or equal to 2. I started with the constraint and transformed it into the desired condition on the LHS then showed that the resulting function on the RHS has a maximum of 2 given the original constraint. Look at the plot also please.

 

[anemone]

Sorry for the late reply to this POTW's thread! I have been real busy with work, and I apologize for MIA for almost two weeks too.

Here is another bright solution of other that I wish to share with the community:

Spoiler

By AM-GM inequality, we have

$2a^5+3\ge 5a^2$ and $2b^3+1\ge 3b^2$

Therefore we get

$2(a^2+b^2) \ge 2(a^5+b^3) \ge 5a^2+3b^2-4$

This simplifies to $3a^2+b^2\le 4$.

Now we need to prove $2a(a+b)\le 3a^2+b^2$, but this is equivalent to $(a-b)^2\ge 0$, hence, the proof follows.