Proving inequalities

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June 15th 2012, 07:19 AM
Kaloda

Proving inequalities

Let a,b, c be non-negative real nos. such that a+b+c=1. Prove that a²+b²+c²+sqrt(12abc)<=1.

Please help me with this one. I don't really have an idea how to prove inequalities. TY.
June 15th 2012, 12:41 PM
richard1234

Re: Proving inequalities

Given $a+b+c = 1$ , squaring both sides yields $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 1$ .

Here, if we can show that $2ab + 2bc + 2ca \ge \sqrt{12abc}$ or $ab + bc + ca \ge \sqrt{3abc}$ , we are done.

Squaring both sides of this inequality, $a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a+b+c) \ge 3abc \Rightarrow a^2b^2 + b^2c^2 + c^2a^2 \ge abc$ (by replacing a+b+c with 1). This is what we want to prove.

By the Cauchy-Schwarz inequality, $(a^2b^2 + b^2c^2 + c^2a^2)(1+1+1) \ge (ab + bc + ca)^2$ . Simplifying, we obtain

$2(a^2b^2 + b^2c^2 + c^2a^2) \ge 2abc(a+b+c) \Rightarrow a^2b^2 + b^2c^2 + c^2a^2 \ge abc$ , which is exactly what we wanted to prove. Hence we're done.
June 16th 2012, 07:10 AM
Kaloda

Re: Proving inequalities

Thank you. But is there another way of proving this without using that Cauchy-Schwarz inequality because I think we didn't tackled it in our class discussion yet.
Or can you give me the brief proof of that inequality without using calculus because I don't have a single knowledge in calculus. Please. TY very MUCH.
June 16th 2012, 08:31 AM
simamura

Re: Proving inequalities

To prove inequality $(a^2b^2+a^2c^2+b^2c^2)(1+1+1) \geq (ab+ac+bc)^2$
consider quadratic function
$(abx+1)^2+(acx+1)^2+(bcx+1)^2$
Since it is sum of squares than it is never negative, therefore its discriminant is never positive: $D \leq 0$ .

Now expand function (recall that $(x+y)^2=x^2+2xy+y^2$ ): $a^2b^2x^2+2abx+1+a^2c^2x^2+2acx+1+b^2c^2x^2+2bcx+1$
or
$(a^2b^2+a^2c^2+b^2c^2)x^2+2(ab+ac+bc)x+1+1+1$

Find discriminant: $D=(ab+ac+bc)^2- (1+1+1)(a^2b^2+a^2c^2+b^2c^2)$
Also, $D \leq 0$
Therefore
$(ab+ac+bc)^2- (1+1+1)(a^2b^2+a^2c^2+b^2c^2) \leq 0$
Or

$(1+1+1)(a^2b^2+a^2c^2+b^2c^2) \geq (ab+ac+bc)^2$

As you see no calculus needed.
June 16th 2012, 09:42 AM
richard1234

Re: Proving inequalities

Cauchy-Schwarz Inequality - AoPSWiki
June 16th 2012, 10:59 AM
Kaloda

Re: Proving inequalities

Why does the discriminant is non-positive? What I knew is that the root of a quadratic function will become imaginary if the discriminant is negative.

Do you know any OTHER WAY or SOLUTION to approach this problem? I think your solutions combined is very long and complicated. But thank you very very much for your replies.

Sorry for my ignorance in higher math. I'm usually good at math but not on this one. It's my first time to solve a problem like this.
June 16th 2012, 12:36 PM
richard1234

Re: Proving inequalities

That's the way inequality problems are; usually you have to use some inequality such as Cauchy-Schwarz, AM-GM, Jensen's, rearrangement inequality, or Muirhead's. Usually you don't see such inequality problems until you take exams such as the USAMO or IMO, which are pretty difficult.

Also, do you see how I got from $(a^2b^2 + b^2c^2 + c^2a^2)(1+1+1) \ge (ab + bc + ca)^2$ to $a^2b^2 + b^2c^2 + c^2a^2 \ge abc$ ? Sorry I skipped a few steps...but if you expand the RHS you should obtain the same thing...
June 16th 2012, 08:21 PM
Kaloda

Re: Proving inequalities

Yea. I think I did traced how you came into that equation.
June 16th 2012, 08:37 PM
richard1234

Re: Proving inequalities

Okay, great. Yeah, the $(a^2b^2 + b^2c^2 + c^2a^2)(1+1+1) \ge (ab + bc + ca)^2$ is essentially the Cauchy-Schwarz inequality.

In general, the Cauchy-Schwarz inequality says that, for non-negative numbers $a_1, a_2, ..., a_n$ and $b_1, b_2, ..., b_n$ , that

$(a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_2^2 + ... + b_n^2) \ge (a_1b_1 + a_2b_2 + ... + a_nb_n)^2$

Equality occurs if and only if $a_i$ is a constant multiple of $b_i$ , that is, $a_i = kb_i$ for all i. It can be generalized even further to complex numbers and vectors.