(A new question of the week)
We have had a number of challenging questions about inequalities recently. I want to show one of those here, because it involved a useful discussion about how to prove them.
A proof problem
Here is the question:
Q) Show that a^4 + b^4 >= (a^3)b + (b^3)a for all real a and b.
My solution:
1) (a^4) – (a^3)b >= (b^3)a – (b^4) … Subtracting (a^3)b and (b^4) on both sides
(a^3)(a – b) >= (b^3)(a – b)
a^3 >= b^3
a >= b ……. [I]2) (b^4) – (a^3)b >= a(b^3) – a^4
By further solving we will get b >= a ….. [II]
By [I] and [II] we get that a = b so the inequality in the question will be proved.
Is this correct? The inequality in the question will always be equal but never will a^4 + b^4 be greater according to my solution. Is there any other way to solve the question?
Arsh has done something many algebra students do: treating a “show” or “prove” problem as if it were a “solve” problem, starting with the equation or inequality. But his “solution” is that a and b are equal, when this is supposed to be true for all pairs of real numbers. Something certainly seems odd.
Doctor Rick answered, pointing out the error in the overall approach, and also a logical error that explains the contradictory conclusion:
Hi, Arsh. There are several errors in your work.
Briefly, if your work concludes that a = b, then you aren’t proving anything is true “for all a and b”. At most you’re proving that a^4 + b^4 = (a^3)b + (b^3)a if and only if a = b.
One error you made, though, is that you divided an inequality by (a – b) on both sides without considering whether (a – b) is positive or negative (or zero).
Correcting the route
Arsh answered, reworking his attempt without correcting the structural error:
I tried the question again and found an answer that was satisfying to me, though I don’t know if it’s really correct.
My solution:
a^4 + b^4 >= (a^3)b + (b^3)a
a^4 – (a^3)b – (b^3)a + b^4 >= 0
(a^3)(a – b) – (b^3)(a – b) >= 0
(a^3 – b^3)(a – b) >= 0
(a – b)(a^2 + ab + b^2) (a – b) >= 0
(a – b)^2 (a^2 + ab + b^2) >= 0.This would be always be >=0 as there are whole squares.
Is this correct? I have some confusion with the term ‘ab’; will that make problems in this solution, since it can be negative?
He has very nicely corrected the error of dividing by a possibly negative number in an inequality, by changing it to factoring. But there is work to do. Doctor Rick replied, first clarifying the structural issue (but setting it aside for later), and then suggesting a fix for the technical issue:
Thanks for your response!
First, what you have written is not a proof, because it starts with the conclusion. However, it is a good exploration for a proof: in effect, you are exploring a river to find its source. Once you’ve found that, you can “float” down the river; that’s the proof.
The part remaining is to prove that a^2 + ab + b^2 ≥ 0 for all real a and b. We can’t ignore the ab term, since it will be negative if a and b have opposite signs.
There may be a more elegant way to prove this inequality as a starting point, but one thing I can suggest is to consider two cases. If a and b have the same sign (or one or both is zero), it’s obvious that a^2 + ab + b^2 ≥ 0. Now consider the case in which a and b have opposite signs, so that ab < 0. Can you rewrite a^2 + ab + b^2 as a square minus ab (or a positive multiple of ab)?
Arsh replied:
Ok according to your words
(a – b)²(a + b)² – ab >= 0
(a – b)²(a + b)² >= ab
on further solving I get –
(a² – b²)² >= ab.As ab < 0, this inequality proves the inequality given in the question. As you said that this is a good exploration for a proof, so is there any other way to prove? Because I have done some more questions like this taking the conclusion inequality and then getting the result in the form that a² >= 0, where a stands for any terms that were given in the question.
Doctor Rick corrected a significant mistake in the algebra (did you catch it?):
You said, (a – b)²(a + b)² – ab >= 0 .
Shouldn’t this be (a – b)² ((a + b)² – ab)?
If you hadn’t made a mistake in that first line, the last line would have been valid. Try again with the correct starting point.
Reversing the flow
We’ll get that fixed in a moment. First, it’s time to look at the big picture:
Do you understand that a correct proof cannot start with the thing to be proved? That’s important! However, all that is needed to make a valid proof in this case is to reverse the steps, so that you end up with the thing to be proved. This is acceptable when every step is reversible. That is, if line (2) not only follows from line (1) but is equivalent to it, then line (1) equally follows from line (2).
Here’s how your exploration can be reworked into a proof in correct form. I’ll leave out the first part.
To prove: a^4 + b^4 >= (a^3)b + (b^3)a for all real a and b.
First, we prove that a^2 + ab + b^2 >= 0 for all real a and b:
If a and b have same sign or zero, then sum of 3 non-negative terms is non-negative.
If a and b have opposite signs, ab < 0, and …
(part to be filled in)
Second, (a – b)^2 >= 0
Thus (a – b)^2 (a^2 + ab + b^2) >= 0
(a – b)(a – b)(a^2 + ab + b^2) >= 0
(a – b)(a^3 – b^3) >= 0
a^4 – ab^3 – a^3 b + b^4 >= 0
a^4 + b^4 >= (b^3)a + (a^3)b
Which was to be proved.Now, while writing this up, I realized that we can actually prove (a – b)(a^3 – b^3) >= 0 more easily, if we can make use of the concept of an increasing function (generally introduced in calculus). This may not be available to you, so probably what we are doing is the sort of proof you are expected to write.
Arsh didn’t show his final work, so let’s fill it in here …
We want to show that \(a^2+ab+b^2\ge0\) when \(ab<0\). As Arsh saw, \(a^2+ab+b^2=(a+b)^2-ab>0\). And that’s all we need for this part of the proof.
I have previously discussed the exploratory phase of a proof in the posts What Role Should a Figure Play in a Proof? and Different Ways to Prove a Trigonometric Identity.