Please continue my path!

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

Given that aa, bb, cc are positive real numbers such that a2+b2+c2+2abc=1a^2+b^2+c^2+2abc=1. Prove that: a2b2+b2c2+c2a212a2b2c2a^2b^2+b^2c^2+c^2a^2 \ge 12a^2b^2c^2.

I've done a few:

a2b2+b2c2+c2a212a2b2c2a^2b^2+b^2c^2+c^2a^2 \ge 12a^2b^2c^2 1a2+1b2+1c212\Longleftrightarrow \frac { 1 }{ a^{ 2 } } +\dfrac { 1 }{ b^ 2 } +\frac { 1 }{ c^ 2 } \ge12

This can be proved if : 9a2+b2+c212 \dfrac{9}{a^2+b^2+c^2} \ge 12 a2+b2+c234\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 a2+b2+c2+2abc=1a^2+b^2+c^2+2abc=1.

Can you all please help me?

Note by Trung Đặng Đoàn Đức
6 years ago

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1 vote

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Well, 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 a2+b2+c234a^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-

a2+b2+c2=12abca^2+b^2+c^2 = 1-2abc

Using AM-GM inequality we get that

a2+b2+c23(abc)2/3a^2+b^2+c^2 \geq 3(abc)^{2/3}

(12abc)327(abc)2\Rightarrow (1-2abc)^3 \geq 27(abc)^2

Substituting u=abcu=abc, then we have

(12u)3=18u36u+12u227u2(1-2u)^3 = 1-8u^3-6u+12u^2 \geq 27u^2 8u3+15u2+6u10\Rightarrow 8u^3+15u^2+6u-1 \leq 0

Let f(u)=8u3+15u2+6u1f(u) = 8u^3+15u^2+6u-1. It's easy to see that f(1)=0f(-1)=0. Thus, after factorizing out u+1u+1, we are left with f(u)=(u+1)(8u2+7u1)f(u)=(u+1)(8u^2+7u-1). And hence f(u)=(u+1)2(8u1)f(u)=(u+1)^2(8u-1).

But we need to find solutions for f(u)0(u+1)2(8u1)0f(u)\leq 0 \Rightarrow (u+1)^2(8u-1) \leq 0.

This gives us u18u \leq \frac{1}{8}. And thus,

abc18\boxed{abc \leq \frac{1}{8}}

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

1a2+1b2+1c23(abc)2/33(8)2/312\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

a2b2+b2c2+c2a212a2b2c2a^2b^2+b^2c^2+c^2a^2 \geq 12a^2b^2c^2

P.S. - Since abc18abc \leq \frac{1}{8}, we have a2+b2+c212(18)=34a^2+b^2+c^2 \geq 1-2\left(\frac{1}{8}\right) = \frac{3}{4}

Kishlaya Jaiswal - 6 years ago

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Thanks :) Learned some tips too :v

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 :)

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Ok, first thing to keep in mind is that you should be familiar with the very basic inequalities like

  • AM-GM Inequality
  • Cauchy-Schwarz Inequality
  • T2's Lemma
  • Power-Mean Inequality
  • Rearrangement Inequality
  • Jensen's Inequality
  • Schur's Inequality

(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 -

  1. If a,b,cR+a,b,c \in \mathbb{R}^+, prove that ab+c+bc+a+ca+b32\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{3}{2}

  2. If aa and bb are positive real numbers such that a+b=1a+b=1. Prove that aabb+abba1a^ab^b+a^bb^a \leq 1

  3. Let a,b,ca,b,c be positive reals such that 1a2+b2+1+1b2+c2+1+1c2+a2+11\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+ca3ab+bc+ca \leq 3

  4. Prove the AM-GM Inequality

BONUS : Try to prove inequality #1 with as many methods as you can.

Kishlaya Jaiswal - 6 years ago

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Very curious to know the answer but too tuff if you find answers post I will also try

Nithin Raju - 6 years ago

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