<s>[INST] Prove the following inequality for any positive real number $x,y$ Prove the following inequality for any positive real number $x,y$
$xy^3 \leq \frac14x^4 + \frac34y^4$
I have tried subtracting $y^4$ from both sides to get:
$xy^3 - y^4 \leq \frac14x^4 - \frac14y^4$
$y^3(x - y) \leq \frac14(x - y)(x + y)(x^2 + y^2)$
$y^3 \leq \frac14(x + y)(x^2 + y^2)$
$0 \leq (x + y)(x^2 + y^2) - 4y^3$
$0 \leq x^3 + x^2y +xy^2 - 3y^3$
But I don't know where to go from here since I couldn't find a way to conveniently factor this. I have tried completing the square from this point, but I couldn't see any obvious way to prove the right side to be greater than zero.
I also tried completing the square from the beginning to get:
$0 \leq x^4 + 3y^4 - 4xy^3$
$0 \leq x^4 + 3y^2(y - \frac23x)^2 -\frac43x^2y^2$
But yet again I couldn't figure out where to go from here. Let me know what I'm doing wrong or what other methods I can try. Thanks!
 [/INST] We need to prove that
$$x^4-4xy^3+3y^4\geq0$$ or
$$x^4-2x^3y+x^2y^2+2x^3y-4x^2y^2+2xy^3+3x^2y^2-6xy^3+3y^4\geq0$$ or
$$(x-y)^2(x^2+2xy+3y^2)\geq0,$$ which is true even for all reals $x$ and $y$.
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