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.
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 $x,\,y$ ?
anemone said:Ah, I'm sorry for posting another challenge with a typo---again...sorry...
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
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.
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.
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 $\frac{1}{z^4}$.
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.
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.