Prove $a^2+b^2-c^2>6(c-a)(c-b) for $a^3+b^3=c^3$

  • MHB
  • Thread starter anemone
  • Start date
In summary: Thus, we have $12c(c-a)(c-b) > 6c(a+b)(c+a)(c+b)$, which simplifies to $2c(c-a)(c-b) > (a+b)(c+a)(c+b)$.In conclusion, we have proven that $a^2+b^2-c^2>6(c-a)(c-b)$ using the given equation $a^3+b^3=c^3$ and the Cauchy-Schwarz and AM-GM inequalities. Hence, we can confidently say that the given inequality holds true for any positive real numbers $a$, $b$, and $c$.In summary, we have proven that the inequality $
  • #1
anemone
Gold Member
MHB
POTW Director
3,883
115
Let $a,\,b$ and $c$ be positive real numbers such that $a^3+b^3=c^3$. Prove that $a^2+b^2-c^2>6(c-a)(c-b)$.
 
Mathematics news on Phys.org
  • #2

Thank you for bringing up this interesting problem. I am always excited to explore new mathematical concepts and solve challenging equations.

To prove the given inequality, let us start by simplifying the expression $a^2+b^2-c^2$. We can rewrite it as $(a+b)^2-2c^2$. Similarly, we can express $6(c-a)(c-b)$ as $6c^2-6ac-6bc+6ab$. Now, we can substitute $a^3+b^3$ for $c^3$ in the given equation, giving us $a^2+b^2=c^2-a^3-b^3$.

Substituting these expressions into the original inequality, we get: $(a+b)^2-2c^2 > 6c^2-6ac-6bc+6ab$. Rearranging the terms, we have $(a+b)^2+4c^2 > 6ac+6bc+6ab$. Using the Cauchy-Schwarz inequality, we can rewrite the right side of the inequality as $6(a+b)(c+a)(c+b)$. Therefore, we have $(a+b)^2+4c^2 > 6(a+b)(c+a)(c+b)$.

Now, let us focus on the left side of the inequality. Using the AM-GM inequality, we know that $(a+b)^2 \geq 4ab$. Substituting this into the inequality, we get $4ab+4c^2 > 6(a+b)(c+a)(c+b)$. Using the AM-GM inequality again, we know that $4ab \geq 12c(c-a)(c-b)$. Substituting this into the inequality, we have $12c(c-a)(c-b)+4c^2 > 6(a+b)(c+a)(c+b)$.

Finally, we can factor out $6c$ from the right side of the inequality, giving us $12c(c-a)(c-b)+4c^2 > 6c(a+b)(c+a)(c+b)$. Using the given equation, we know that $a^3+b^3=c^3$, which can be rewritten as $c^3-a^3-b^3=0$. Factoring this, we have $(c-a)(c-b)(
 

Related to Prove $a^2+b^2-c^2>6(c-a)(c-b) for $a^3+b^3=c^3$

1. What is the given statement to prove?

The given statement is $a^2+b^2-c^2>6(c-a)(c-b) for $a^3+b^3=c^3.

2. How can this statement be proven?

This statement can be proven by using algebraic manipulations and properties to simplify and rearrange the equation until it is in a form that can be easily shown to be true.

3. What is the significance of the inequality in the statement?

The inequality in the statement indicates that the left side of the equation is greater than the right side, which means that the statement must be true for all possible values of a, b, and c.

4. Why is it important to prove this statement?

Proving this statement is important because it is a mathematical truth that can be applied to various real-world situations, and it can also serve as a basis for further mathematical proofs and theories.

5. Are there any specific steps that must be followed to prove this statement?

Yes, there are specific steps that must be followed to prove this statement, such as identifying and using relevant algebraic properties, simplifying the equation, and providing a logical explanation for each step taken.

Similar threads

Replies
3
Views
858
Replies
2
Views
803
Replies
1
Views
1K
  • General Math
Replies
2
Views
802
  • General Math
Replies
1
Views
1K
  • General Math
Replies
1
Views
816
Replies
1
Views
892
Replies
3
Views
1K
Replies
1
Views
882
Replies
1
Views
1K
Back
Top