Disguised Inequality

Algebra Level 4

$(a+b+c)(a^2+b^2+c^2)+3abc\geq N(a+b)(a+c)(b+c)$

If $a,b,c$ and $N$ are positive reals such that the maximum value of $N$ which satisfies the inequality above can be expressed in the form $\frac{m}{n}$ , where $m$ and $n$ are coprime positive integers, find the value of $m+n$ .

The answer is 5.

3 solutions

Calvin Lin Staff
Feb 20, 2016

We suspect that equality occurs when $a = b = c$ , which yield a value of $N = \frac{3}{2}$ . Thus, we want to show that

$2 [ (a+b+c) ( a^2 + b^2 + c^2 ) + 3abc ] \geq 3 (a+b)(b+c)(c+a).$

Expanding both sides and subtracting terms, this is equivalent to

$2a^3 + 2b^3 + 2c^3 \geq a^2b + b^2 c + c^2 a + a^2 c + b^2 a + c^2b,$

which is obviously true.

How is the last inequality obvious (it's not obvious to me)?

Deeparaj Bhat - 5 years, 3 months ago

See Applying the Arithmetic Mean Geometric Mean Inequality .

Calvin Lin Staff - 5 years, 3 months ago

Zk Lin
Feb 14, 2016

Note that

L.H.S. $=a^{3}+b^{3}+c^{3}+3abc +\displaystyle \sum_{sym} a^{2}b$

$\geq 2 \displaystyle \sum_{sym} a^{2}b$ by Schur's inequality

$\geq 12abc$ by AM-GM inequality.

Equality is attained when $a=b=c$ .

R.H.S. $=2Nabc+N\displaystyle \sum_{sym} a^{2}b$

$\geq 2Nabc+ 6Nabc$ by AM-GM inequality

$=8Nabc$

Equality is attained when $a=b=c$ .

Since L.H.S. $\geq$ R.H.S. $\geq 8Nabc$ ,

we have L.H.S. $\geq 8Nabc$ .

However, we have already proven that the minimum of L.H.S. is attained at $12abc$ .

Therefore $8N=12$ , which simplifies to $N=\frac{3}{2}$ .

This gives $m=3,n=2$ , hence the answer $3+2=\boxed{5}$ .

Moderator note:

This solution is non-sensical. Do you see why?

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin I am not quite sure, but I think it has to do with

Since L.H.S. $\geq$ R.H.S. $\geq 8Nabc$ ,

we have L.H.S. $\geq 8Nabc$ .

Can you explain why this approach is wrong?

ZK LIn - 5 years, 3 months ago

Depending on your interpretation / thoughts as you were writing it up:

Even if it's true that $LHS \geq 8 abc$ , how does that imply that $LHS \geq \frac{3}{2} (a+b)(b+c)(c+a)$ ? All that we have is $\frac{3}{2} (a+b)(b+c)(c+a) \geq 8 abc$ .
In essence, we want $\frac{LHS}{ (a+b)(b+c)(c+a) } \geq N$ . To continue your approach (where we treated the terms independently, we have to minimize LHS and maximize $(a+b)(b+c)(c+a)$ .
It is useful to list inequalities as "Want to show" or "It is equaivalent to show that" VS "We know that". It can be easy to mix these up when we have a long list of statements written down.

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin – @Calvin Lin In order to try to understand what has gone wrong, I go over my thought process while solving this question and try to reconstruct the whole argument backwards, starting from the fact that L.H.S. $\geq 12abc$ (We know this is true from Schur's + AM-GM). Therefore, if L.H.S. $\geq 8Nabc$ , then such greatest $N$ is $\frac{12}{8}=\frac{3}{2}$ .

From L.H.S. $\geq 8Nabc$ , we have L.H.S $\geq$ R.H.S. $\geq 8Nabc$ . The first part of the inequality is given in the question while R.H.S. $\geq 8Nabc$ is separately proven above .

Therefore, we have L.H.S. $\geq$ R.H.S , or

$(a+b+c)(a^2+b^2+c^2)+3abc\geq N(a+b)(a+c)(b+c)$

substitute $N=\frac{3}{2}$ , the inequality should in theory, hold, and we note that such $N$ is the maximum permitted.

I think the bolded part is fallacious. I committed a logical fallacy while assuming L.H.S $\geq$ R.H.S. $\geq 8Nabc$ implies L.H.S. $\geq$ R.H.S for all $N$ in the backward proof when in actual fact, the sign of the inequality might flip depending on the value of $N$ . In other words, I cannot assume that the inequality holds, find the maximum $N=\frac{3}{2}$ and claim that I have solved the problem.

Similarly, in my original proof (forward proof), I cannot infer from L.H.S $\geq$ R.H.S. $\geq$ $8Nabc$ that L.H.S. $\geq 8Nabc$ . Who knows for certain values of $N$ , the inequality flips to R.H.S. $\geq 8Nabc \geq$ L.H.S.

Now that I have noticed this, this fallacy is indeed quite dangerous. While typing up the solution, it didn't even cross my mind that I have assumed things I shouldn't.

What do you think about it? Do I correctly identify the flaw in my proof?

ZK LIn - 5 years, 3 months ago

@Zk Lin – Right, we have shown that $LHS \geq 12 abc$ and $RHS \times N \geq 8abc \times N$ . This however does not imply that $LHS \geq \frac{3}{2} RHS$ , since we need to bound the RHS from above, instead of from below.

This is a pretty common mistake made when manipulating such "fractional" inequalities, and not being careful about how we're multiplying / comparing them. That's one reason why Titu's lemma is so powerful, because it helps you make the correct division step.

Calvin Lin Staff - 5 years, 3 months ago

Chew-Seong Cheong
Jan 6, 2015

$\color{#D61F06}{\text{Sorry, this is incorrect.}}$

We note that:

$(a+b)(b+c)(c+a) = (a+b+c)(ab+bc+ca)-abc$

Using Cauchy-Schwarz inequality we have:

$a+b+c \ge 3(abc)^{\frac{1}{3}}\quad \text{and} \quad ab+bc+ca \ge 3(abc)^{\frac{2}{3}}$

$\Rightarrow (a+b+c)(ab+bc+ca) \ge 9abc\quad \Rightarrow (a+b)(b+c)(c+a) \ge 8abc$

We also note that:

$\begin{aligned} (ab+bc+ca)^2 & \le (a^2+b^2+c^2)(a^2+b^2+c^2) \\ a^2+b^2+c^2 & \ge ab+bc+ca \\ (a+b+c)(a^2+b^2+c^2) & \ge (a+b+c)(ab+bc+ca) \\(a+b+c)(a^2+b^2+c^2) & \ge (a+b)(b+c)(c+a) + abc \\ (a+b+c)(a^2+b^2+c^2) + 3abc & \ge (a+b)(b+c)(c+a) + 4abc \\ (a+b+c)(a^2+b^2+c^2) + 3abc & \ge (a+b)(b+c)(c+a) + \frac {4}{8} (a+b)(b+c)(c+a) \\ (a+b+c)(a^2+b^2+c^2) + 3abc & \ge \frac {3}{2} (a+b)(b+c)(c+a) \end{aligned}$

$\Rightarrow m = 3$ , $n=2$ and $m+n = \boxed{5}$

I don't understand how does the 4th last line imply the 3rd last line. isn't 1/8(a+b)(b+c)(c+a) more than abc?

Wuu Yyiizzhhoouu - 5 years, 4 months ago

You are right. Let me redo the problem.

Chew-Seong Cheong - 5 years, 4 months ago

thank you so much!

Wuu Yyiizzhhoouu - 5 years, 4 months ago

You should also write the equality cases as each may have been different from one another.

Reineir Duran - 5 years, 4 months ago

Disguised Inequality

The answer is 5.

3 solutions

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