Proof using AM-GM inequality

Question

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?

Answer 1

Because by AM-GM $$3y^4+x^4\geq4\sqrt[4]{\left(y^4\right)^3x^4}=4|xy^3|\geq4xy^3$$ and $$x^4+y^4-x^3y-xy^3=(x-y)(x^3-y^3)=(x-y)^2(x^2+xy+y^2)\geq$$ $$\geq(x-y)^2\left(2\sqrt{x^2y^2}+xy\right)=(x-y)^2\left(2|xy|+xy\right)\geq0.$$

Answer 2

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

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

(ii) follows from (i):

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