Homework is given to a child who is not ready for a math competition - (2). - Thành Đạt Lê

My teacher gave me this problem:

$a$, $b$ and $c$ are three positives such that $abc = 1$. Prove that

$\large M = \dfrac{1}{a^2 - a + 1} + \dfrac{1}{b^2 - b + 1} + \dfrac{1}{c^2 - c + 1} \le 3$

Here's how I solve it.

$\large \begin{aligned} M &= \dfrac{1}{a^2 - a + 1} + \dfrac{1}{b^2 - b + 1} + \dfrac{1}{c^2 - c + 1}\\ &= \left(a + 1 - \dfrac{a^3}{a^2 - a + 1}\right) + \left(b + 1 - \dfrac{b^3}{b^2 - b + 1}\right) + \left(c + 1 - \dfrac{c^3}{c^2 - c + 1}\right) && m^3 + 1 = (m + 1)(m^2 - m + 1)\\ &= (a + b + c) - \left(\dfrac{a^2}{bc + a - 1} + \dfrac{b^2}{ca + b - 1} + \dfrac{c^2}{ab + c - 1}\right) + 3 && abc = 1\\ &\le (a + b + c) - \dfrac{(a + b + c)^2}{(ab + bc + ca) + (a + b + c) - 3} + 3 && \text{Sedrakyan's inequality, Engel’s form or Titu’s lemma}\\ &\le (a + b + c) - \dfrac{(a + b + c)^2}{\dfrac{(a + b + c)^2}{3} + (a + b + c) - 3} + 3 && ab + bc + ca \le a^2 + b^2 + c^2\\ \end{aligned}$

(Fun fact, the Sedrakyan's inequality is called the Schwarz inequality here.)

Let $a + b + c = n$ , we have that $a + b + c \ge 3\sqrt[3]{abc} = 3$ and $M \le n - \dfrac{3n^2}{n^2 + 3n - 9} + 3$ .

The problem becomes:

Calculate the maximum value of the following expression if $n \ge 3$ .

$\large M' = n - \dfrac{3n^2}{n^2 + 3n - 9} + 3$

So if you can:

solve the next part of the solution
(prove that there is no maximum value of $M'$ and) solve the original problem in a faster, more convenient way

then feel free to write them in the comments' section. Please help me!

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

Well, the expression is in one variable 'n' and also, we know from the given condition of abc=1 that a+b+c = n is atleast 3......(AM-GM)........Now, by simple first derivative,we see that the expression in M is always increasing for n>=3..........And hence, the maximum value of M is 3 which occurs when n = 3 or a+b+c=3 or a=b=c=1.........!!! Here is a graph supporting the answer..........

Also, my approach was on the same lines as yours........I usually try to reduce inequalities to a single variable, just like you have done.....!!

Aaghaz Mahajan - 2 years, 5 months ago

I am really sorry but I am just in grade 9 so I can't use derivatives, I hope that you can come up a solution completely in pure algebra.

Thành Đạt Lê - 2 years, 5 months ago

Now, by simple first derivative,we see that the expression in M is always increasing for n>=3

Care to elaborate? I don't see the connection between the Desmos graph and the inequality posed.

Pi Han Goh - 2 years, 5 months ago

Here Sir, I have added the inequality of the derivative, showing the region where the function is increasing...........Hope it is clear now......:)

M is always increasing for n>=3..........And hence, the maximum value of M is 3 which occurs when n = 3 or a+b+c=3 or a=b=c=1.........!!!

Your first statement claims that $M$ has no upper bound. But your second statement says it has an upper bound of 3. How is that possible?

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$