Proving Inequalities Involving Positive Real Numbers

  • Thread starter Thread starter Sam Morse
  • Start date Start date
AI Thread Summary
The inequality a^3 + b^3 + c^3 ≥ a^2b + b^2c + c^2a for positive real numbers a, b, and c is under discussion. The initial approach involved assuming a ≥ b ≥ c > 0 and attempting to rearrange the inequality. A suggestion was made to simplify the problem by dividing by abc, leading to a derived inequality a^2 + b^2 + c^2 ≥ ab + ac + bc. The discussion highlights the importance of proper factorization to prove the inequality, with a reference to an external source for further assistance. Ultimately, the goal is to establish the validity of the inequality through algebraic manipulation.
Sam Morse
Messages
15
Reaction score
0

Homework Statement



Let a,b,c be positive real numbers. Prove that:

a3+b3+c3≥a2b+b2c+c2a

Homework Equations





The Attempt at a Solution



I assumed that a≥b≥c>0 following which I shifted the left side of this inequality to the right side giving

a3+b3+c3-(a2b+b2c+c2a)≥0

How do I do the required factorisation ... ?
 
Physics news on Phys.org
Are you sure that there is an easy way to factorize this?

I would expect that this can be reduced to a problem similar to a/b + b/c + c/a > 3.
 
I see a way to solve it, though it is a bit messy. First divide the entire equation by abc .
A bit of algebra gets a^3+b^3+c^3 \geq ab + bc+ ac. We can assume that a,b, and c are all greater than 1 since if (a,b,c) satisfies the inequality, then so does (ax,bx,cx) for any positive x. Using the fact that x^3 \geq x^2 if x>1, it is therefore sufficient to show that
a^2+b^2+c^2 \geq ab + bc +ac

Get everything over to the left hand side and multiply by 2. You will see that it factors nicely.

Edit: Oops, disregard what I said, I made a mistake in my algebra.
 
Last edited:
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top