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

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The discussion focuses on proving the inequality $\frac{1}{x^4}+\frac{1}{4x^3y}+\frac{1}{6x^2y^2}+\frac{1}{4xy^3}+\frac{1}{y^4} \geq \frac{128}{3(x+y)^4}$ for positive real numbers $x$ and $y$. There was a clarification regarding a typo in the original post, specifically the exponent of $y$ in one term. Participants acknowledged the importance of accuracy in problem posting while appreciating the effort involved in creating such challenges. The conversation highlights the collaborative nature of problem-solving in the forum. Overall, the thread emphasizes both the mathematical challenge and the community's supportive environment.
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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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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 $x,\,y$ ?
 
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.
 
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)
 
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
 
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!
 
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