# Proving inequalities

• Jun 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.
• Jun 15th 2012, 12:41 PM
richard1234
Re: Proving inequalities
Given $\displaystyle a+b+c = 1$, squaring both sides yields $\displaystyle a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 1$.

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

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

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

$\displaystyle 2(a^2b^2 + b^2c^2 + c^2a^2) \ge 2abc(a+b+c) \Rightarrow a^2b^2 + b^2c^2 + c^2a^2 \ge abc$, which is exactly what we wanted to prove. Hence we're done.
• Jun 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.
• Jun 16th 2012, 08:31 AM
simamura
Re: Proving inequalities
To prove inequality $\displaystyle (a^2b^2+a^2c^2+b^2c^2)(1+1+1) \geq (ab+ac+bc)^2$
$\displaystyle (abx+1)^2+(acx+1)^2+(bcx+1)^2$
Since it is sum of squares than it is never negative, therefore its discriminant is never positive: $\displaystyle D \leq 0$.

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

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

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

As you see no calculus needed.
• Jun 16th 2012, 09:42 AM
richard1234
Re: Proving inequalities
• Jun 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.
• Jun 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 $\displaystyle (a^2b^2 + b^2c^2 + c^2a^2)(1+1+1) \ge (ab + bc + ca)^2$ to $\displaystyle a^2b^2 + b^2c^2 + c^2a^2 \ge abc$? Sorry I skipped a few steps...but if you expand the RHS you should obtain the same thing...
• Jun 16th 2012, 08:21 PM
Kaloda
Re: Proving inequalities
Yea. I think I did traced how you came into that equation.
• Jun 16th 2012, 08:37 PM
richard1234
Re: Proving inequalities
Okay, great. Yeah, the $\displaystyle (a^2b^2 + b^2c^2 + c^2a^2)(1+1+1) \ge (ab + bc + ca)^2$ is essentially the Cauchy-Schwarz inequality.

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

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

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