Prove Inequality: IMO $\frac{1}{x^4}+\cdots \geq \frac{128}{3(x+y)^4}$

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In summary, there was a typo in the original post which has been corrected. The challenge is to prove that the given inequality holds true for positive real numbers $x$ and $y$.
  • #1
anemone
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Prove $\dfrac{1}{x^4}+\dfrac{1}{4x^3y} + \dfrac{1}{6x^2y^2}+ \dfrac{1}{4xy^3}+ \dfrac{1}{y^4} ≥ \dfrac{128}{3(x+y)^4}$, given $x,\,y$ are positive real numbers.
 
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  • #2
anemone said:
Prove $\dfrac{1}{x^4}+\dfrac{1}{4x^3y} + \dfrac{1}{6x^2y^2}+ \dfrac{1}{4xy^4}+ \dfrac{1}{y^4} ≥ \dfrac{128}{3(x+y)^4}$, given $x,\,y$ are positive real numbers.

Do you mean $\dfrac{1}{x^4}+\dfrac{1}{4x^3y} + \dfrac{1}{6x^2y^2}+ \dfrac{1}{4xy^3}+ \dfrac{1}{y^4} ≥ \dfrac{128}{3(x+y)^4}$, given $x,\,y$ ?
 
  • #3
RLBrown said:
Do you mean $\dfrac{1}{x^4}+\dfrac{1}{4x^3y} + \dfrac{1}{6x^2y^2}+ \dfrac{1}{4xy^3}+ \dfrac{1}{y^4} ≥ \dfrac{128}{3(x+y)^4}$, given $x,\,y$ ?

Ah, I'm sorry for posting another challenge with a typo---again...sorry RLBrown, your intuition is right, and I will correct my first post to change the exponent of $y$ to 3 in the fourth term on the LHS of the inequality, thanks for letting me know about it.
 
  • #4
anemone said:
Ah, I'm sorry for posting another challenge with a typo---again...sorry...

You know, only those who post challenges are at risk of posting challenges with typos. The only way one can prevent oneself from making mistakes is to not do anything. Given the hundreds of such problems you have tirelessly and diligently posted for our enjoyment here at MHB and the very small number with typos, I would say your track record is excellent. (Yes)
 
  • #5
my solution
by recombinaion and using $AP \geq GP$ we have :
left side :
$(\dfrac{1}{x^4}+\dfrac{1}{y^4})+(\dfrac{1}{4x^3y}+\dfrac{1}{4xy^3})+\dfrac{1}{6x^2y^2}\
\geq \dfrac {2}{x^2y^2}+\dfrac {1}{2x^2y^2}+\dfrac {1}{6x^2y^2}=\dfrac{8}{3x^2y^2}$
right side :
$\dfrac{128}{3(x+y)^4}\leq \dfrac {128}{48x^2y^2}=\dfrac{8}{3x^2y^2}$
and the proof is done
 
  • #6
Albert said:
my solution
by recombinaion and using $AP \geq GP$ we have :
left side :
$(\dfrac{1}{x^4}+\dfrac{1}{y^4})+(\dfrac{1}{4x^3y}+\dfrac{1}{4xy^3})+\dfrac{1}{6x^2y^2}\
\geq \dfrac {2}{x^2y^2}+\dfrac {1}{2x^2y^2}+\dfrac {1}{6x^2y^2}=\dfrac{8}{3x^2y^2}$
right side :
$\dfrac{128}{3(x+y)^4}\leq \dfrac {128}{48x^2y^2}=\dfrac{8}{3x^2y^2}$
and the proof is done

Well done, Albert and thanks for participating!
 

FAQ: Prove Inequality: IMO $\frac{1}{x^4}+\cdots \geq \frac{128}{3(x+y)^4}$

What is the purpose of proving inequalities in mathematics?

Inequalities are used to compare two values or expressions and determine which is larger or smaller. They are important in mathematics as they help to establish relationships between quantities and can be used to solve equations and problems.

What is the significance of the inequality "IMO $\frac{1}{x^4}+\cdots \geq \frac{128}{3(x+y)^4}$"?

This inequality is significant because it was used as a problem in the International Mathematical Olympiad (IMO), which is a prestigious international competition for high school students. Proving this inequality demonstrates a deep understanding of mathematical concepts and is a challenging problem for students to solve.

What does the notation $\frac{1}{x^4}+\cdots$ mean?

The notation $\frac{1}{x^4}+\cdots$ is a shorthand way of representing a series of terms, where the ellipsis (...) indicates that there are more terms in the series. In this particular inequality, the ellipsis represents other terms that follow the first term $\frac{1}{x^4}$, such as $\frac{1}{y^4}$ and $\frac{1}{z^4}$.

How can I prove this inequality?

The proof of this inequality involves using mathematical techniques such as algebra, calculus, and inequality theorems. It also requires a good understanding of the properties of exponents and fractions. It is a difficult problem and may require multiple steps and strategies to solve. It is recommended to seek guidance from a teacher or mentor if attempting to prove this inequality.

Why is it important to include the condition $(x,y>0)$ in the inequality?

The condition $(x,y>0)$ is important because the inequality is not true for all values of $x$ and $y$. For example, if $x=-1$ and $y=1$, the inequality becomes $\frac{1}{(-1)^4}+\frac{1}{1^4} \geq \frac{128}{3(0)^4}$, which is not a valid inequality. By including the condition $(x,y>0)$, we ensure that the inequality is only being compared for positive values of $x$ and $y$, making the inequality valid.

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