Tuesday, April 2, 2024

Algebra+ 05: UNIZOR.COM - Math+ & Problems- Algebra

Notes to a video lecture on http://www.unizor.com

Algebra+ 05

Problem A

Given a system of two equations with three unknown variables x, y and z:
x + y + z = A
x−1 + y−1 + z−1 = A−1
Prove that one of the unknown variables equals to A.

Hint A

System of equations
x + y = p
x · y = q
fully defines a pair of numbers (generally speaking, complex numbers) as solutions to a quadratic equation
X² − p·X + q.
Indeed, if X1 and X2 are the solution of the equation, then, according to the Vieta's Theorem,
X1 + X2 = −(−p) = p and
X1 · X2 = q
(See a lecture Math 4 Teens - Algebra - Quadratic Equations - Lecture on UNIZOR.COM)

The only unresolved issue is:
which unknown variable takes which value from a pair
is it (x=X1,y=X2) or (x=X2,y=X1).

From this follows that, if the following system of equations is given
x + y = a + b
x · y = a · b
then either (x=a,y=b)
or (x=b,y=a).

Proof A

x + y = A − z
x−1 + y−1 = A−1 − z−1
None of the unknown variables can be equal to zero, since each is represented in the second equation in the denominator.
Therefore, we can multiply the second equation by x·y getting
y + x = x·y/A − x·y/z

Using the first equation, substitute x+y into the second getting a system of equations
x + y = A − z
A − z = x·y·(1/A − 1/z)
x + y = A − z
x·y = (A−z)/(1/A − 1/z)
x + y = A − z
x·y = −A·z
x + y = A + (−z)
x·y = A·(−z)
Therefore, either x=A, y=−z
or y=A, x=−z.

Problem B

Prove that
(x + y)4 ≤ 8·x4 + 8·y4

Proof B

Let's start with analysis of this problem.
Assume, this inequality (call it "statement A") is true and make invariant (reversible and equivalent) transformations to it, trying to get to an obviously true statement B.
Then, using the fact that our transformations were invariant, we can say that we can start with obviously true statement B and, using the reverse transformations, derive statement A, that is we will prove that A is true.

Notice that, if we divide both sides of this inequality by y4 and assign t=x/y, we will reduce the number of variables from two to one, which seems to simplify the task.
Dividing by positive y4 is an invariant transformation of an inequality, except a case of y=0. The case of y=0 can be considered separately, and in this case the inequality is obviously true since x4 ≤ 8·x4.

After dividing by y4 and substituting t=x/y the new inequality looks like
(t + 1)4 ≤ 8t4 + 8
which seems to be simpler to prove.

Let's open the parenthesis and bring all items to one side of an inequality - obviously invariant transformation
7t4 − 4t3 − 6t2 − 4t + 7 ≥ 0
Notice that the sum of coefficients of a polynomial on the left is zero. That means that t=1 is a root of this polynomial, that is it's equal to zero for t=1.

Recall the Fundamental Theorem of Algebra (see Math 4 Teens course on UNIZOR.COM, menu items Algebra - Fundamental Theorem of Algebra and its Corollary 1) that states that if x=a is a root of a polynomial P(n)(x) of nth degree, then this polynomial is divisible by x−a, that is
P(n)(x) = (x−a)·Q(n−1)(x)
where Q(n−1)(x) is a polynomial of a degree lower by 1 than P(n)(x).

Therefore, since t=1 is a root of the polynomial of the 4th degree above, we can represent that polynomial as (t−1) multiplied by another polynomial of the 3rd degree.
7t4 − 4t3 − 6t2 − 4t + 7 =
= (t − 1)·(7t3 + 3t2 − 3t − 7)

Consider the polynomial
7t3 + 3t2 − 3t − 7
The sum of its coefficient is zero too. Therefore, we can represent it as a product of (t−1) and a polynomial of the second degree
7t3 + 3t2 − 3t − 7 =
= (t − 1)·(7t2 + 10t + 7)

So, the inequality we have to prove was transformed into this one:
(t − 1)2·(7t2 + 10t + 7) ≥ 0
In this inequality the member (t−1)2 is always greater or equal to zero.
Quadratic polynomial 7t2+10t+7 has discriminant Δ=102−4·7·7=−96, which is negative and, consequently, it has no roots, it's always not equal to zero.
It can only be greater than zero since the coefficient at t2 is positive.
Therefore, this polynomial is always greater than zero.
That concludes the analysis of our problem.

The proof proper is to start from an obviously truthful statement
(t − 1)2·(7t2 + 10t + 7) ≥ 0
and transform it into
(t + 1)4 ≤ 8t4 + 8
Replacing t with x/y (recall, a trivial case y=0 was already checked, so now we assume that y≠0) and multiplying by y4 finishes the proof.

Problem C

Prove the following inequality
x12 − x9 + x4 − x + 1 > 0

Proof C

Consider a polynomial
x12 − x9 + x4 − x
It can be invariantly transformed into
x9·(x3 − 1) + x·(x3 − 1) or
(x3 − 1)·(x9 + x) or
(x3 − 1)·x·(x8 + 1)

This polynomial has only two roots: x=0 and x=1
As easily checked, values outside interval (0,1) are non-negative and inside this interval the values of a polynomial are negative.
Since we are interested in the values of this polynomial +1, the only interval where it's not obvious whether after adding 1 it is positive or not is inside the interval (0,1).

Inside interval (0,1)
x12 − x9 + x4 − x + 1 =
= x12 + x4·(1−x5) + (1−x)

with every item in parenthesis and every other participant in the above expression is positive, which results in a positive value of an entire expression.

End of proof.

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