Vrftjkry

The questions has two parts:

Prove

(i) $ xy^{3} leq frac{1}{4}x^{4} + frac{3}{4}y^{4} $

and

(ii) $ xy^{3} + x^{3}y leq x^{4} + y^{4}$.

Now then, I went about putting both sides of $sqrt{xy} leq frac{1}{2}(x+y)$
to the power of 4 and it left me with

$$-x^{3}y leq frac{1}{4}x^{4} + frac{1}{4}y^{4} + xy^{3} + frac{5}{2}x^{2}y^{2}. $$

Curiously squaring and multiplying $sqrt{xy} leq frac{1}{2}(x+y)$ I've tried merging the results with my other inequality a few times to no avail - I just can't seem to get the signs right and nor can I seem to make the coefficients of the $x^{4}$ and $y^{4}$ different, as they are in (i). I feel like there's something I'm missing. Does anyone see a nice way about this problem?

Because by AM-GM $$3y^4+x^4geq4sqrt[4]{left(y^4right)^3x^4}=4|xy^3|geq4xy^3$$ and
$$x^4+y^4-x^3y-xy^3=(x-y)(x^3-y^3)=(x-y)^2(x^2+xy+y^2)geq$$
$$geq(x-y)^2left(2sqrt{x^2y^2}+xyright)=(x-y)^2left(2|xy|+xyright)geq0.$$

(i) follows from Young's inequality for $p = 4$ and $q = frac43$:

$$xy^3 le |x||y|^3 le frac{|x|^p}{p} + frac{left(|y|^3right)^q}q = frac{|x|^4}4 + frac{left(|y|^3right)^{4/3}}{4/3} = frac{|x|^4+3|y|^4}4 = frac{x^4+3y^4}4$$

(ii) follows from (i):

$$xy^3 + x^3y le frac{x^4+3y^4}4 + frac{3x^4+y^4}4 = x^4+y^4$$