While going through inequality books, I've found a very interesting problem:

Given that \(a\), \(b\), \(c\) are positive real numbers such that \(a^2+b^2+c^2+2abc=1\). Prove that: \(a^2b^2+b^2c^2+c^2a^2 \ge 12a^2b^2c^2\).

I've done a few:

\[a^2b^2+b^2c^2+c^2a^2 \ge 12a^2b^2c^2\] \[\Longleftrightarrow \frac { 1 }{ a^{ 2 } } +\dfrac { 1 }{ b^ 2 } +\frac { 1 }{ c^ 2 } \ge12\]

This can be proved if : \[ \dfrac{9}{a^2+b^2+c^2} \ge 12\] \[\Longleftrightarrow a^2 + b^2 + c^2 \le \dfrac{3}{4} \]

This is where I've stucked, since I cannot think of any connection between it and the fact that \(a^2+b^2+c^2+2abc=1\).

Can you all please help me?

## Comments

Sort by:

TopNewestWell, one thing I would like to suggest is that never try to reverse-engineer an inequality. Most of times you'll end up with something incorrect. As you can see here - you pose that \(a^2+b^2+c^2 \leq \frac{3}{4}\) but infact, the opposite inequality holds true.

EDIT : Although sometimes, you might end up with a trivial solution for an inequality after reverse-engineering it.

Anyways, here's the solution.

First, let's try to deduce some results from the given expression. We have-

\[a^2+b^2+c^2 = 1-2abc\]

Using AM-GM inequality we get that

\[a^2+b^2+c^2 \geq 3(abc)^{2/3}\]

\[\Rightarrow (1-2abc)^3 \geq 27(abc)^2\]

Substituting \(u=abc\), then we have

\[(1-2u)^3 = 1-8u^3-6u+12u^2 \geq 27u^2\] \[\Rightarrow 8u^3+15u^2+6u-1 \leq 0\]

Let \(f(u) = 8u^3+15u^2+6u-1\). It's easy to see that \(f(-1)=0\). Thus, after factorizing out \(u+1\), we are left with \(f(u)=(u+1)(8u^2+7u-1)\). And hence \(f(u)=(u+1)^2(8u-1)\).

But we need to find solutions for \(f(u)\leq 0 \Rightarrow (u+1)^2(8u-1) \leq 0\).

This gives us \(u \leq \frac{1}{8}\). And thus,

\[\boxed{abc \leq \frac{1}{8}}\]

Now, let's try to implement this result. Observe that, by AM-GM inequality we have

\[\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \geq \frac{3}{(abc)^{2/3}} \geq 3(8)^{2/3} \geq 12\]

On rearranging we get our desired result

\[a^2b^2+b^2c^2+c^2a^2 \geq 12a^2b^2c^2\]

P.S. - Since \(abc \leq \frac{1}{8}\), we have \(a^2+b^2+c^2 \geq 1-2\left(\frac{1}{8}\right) = \frac{3}{4}\) – Kishlaya Jaiswal · 1 year, 8 months ago

Log in to reply

By the way, can you give me some tricks or advices to solve inequalities? I usually do reverse-engineering but I've completely changed my mind after this :) – Trung Đặng Đoàn Đức · 1 year, 8 months ago

Log in to reply

(I hope I am not missing out any important ones)

Then must try out examples as many as you can. Here are few to start with -

If \(a,b,c \in \mathbb{R}^+\), prove that \[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{3}{2}\]

If \(a\) and \(b\) are positive real numbers such that \(a+b=1\). Prove that \[a^ab^b+a^bb^a \leq 1\]

Let \(a,b,c\) be positive reals such that \[\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1} \geq 1\]. Prove that \[ab+bc+ca \leq 3\]

Prove the AM-GM Inequality

BONUS : Try to prove inequality #1 with as many methods as you can. – Kishlaya Jaiswal · 1 year, 8 months ago

Log in to reply

Very curious to know the answer but too tuff if you find answers post I will also try – Nithin Raju · 1 year, 8 months ago

Log in to reply