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

[anemone]

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.

 

[RLBrown]

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 $x,\,y$ ?

 

[anemone]

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.

 

[MarkFL]

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)

 

[Albert1]

my solution

Spoiler

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

 

[anemone]

my solution

Spoiler

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!