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

My teacher gave me this problem:

aa, bb and cc are three positives such that abc=1abc = 1. Prove that

M=1a2a+1+1b2b+1+1c2c+13\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.

M=1a2a+1+1b2b+1+1c2c+1=(a+1a3a2a+1)+(b+1b3b2b+1)+(c+1c3c2c+1)m3+1=(m+1)(m2m+1)=(a+b+c)(a2bc+a1+b2ca+b1+c2ab+c1)+3abc=1(a+b+c)(a+b+c)2(ab+bc+ca)+(a+b+c)3+3Sedrakyan’s inequality, Engel’s form or Titu’s lemma(a+b+c)(a+b+c)2(a+b+c)23+(a+b+c)3+3ab+bc+caa2+b2+c2\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=na + b + c = n, we have that a+b+c3abc3=3a + b + c \ge 3\sqrt[3]{abc} = 3 and Mn3n2n2+3n9+3M \le n - \dfrac{3n^2}{n^2 + 3n - 9} + 3.

The problem becomes:

Calculate the maximum value of the following expression if n3n \ge 3.

M=n3n2n2+3n9+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 MM' and) solve the original problem in a faster, more convenient way

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

Note by Thành Đạt Lê
5 months ago

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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 - 5 months ago

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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 - 5 months ago

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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 - 5 months ago

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

Aaghaz Mahajan - 5 months ago

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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 MM has no upper bound. But your second statement says it has an upper bound of 3. How is that possible?

Pi Han Goh - 5 months ago

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