# Proving $4(x^4+y^4+z^4)\geq \sum_{cyc}xy(x+y)^2\geq 4xyz(x+y+z)$, for $x,y,z\geq0$

Trying to get used to olympiad inequalities. Tried AM-GM and did not succed. Please explain. (I'm an eight grader)

Prove that for $$x,y,z\geq 0$$, the following inequality holds true: $$4(x^4+y^4+z^4)\geq xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\geq 4xyz(x+y+z)$$

• AMGM almost certainly works for the first inequality. It also works for the second if you futz around with it for a bit. Commented Feb 24, 2023 at 22:58

Note that the second inequality (right most) is homogeneous, so you can scale by some factor to get $$x+y+z = 1$$. Then go to If $x,y,z>0$ and $x+y+z=1$ Then prove that $xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\geq 4xyz$.

Let's now focus on proving $$4(x^4+y^4+z^4)\geq xy(x+y)^2+yz(y+z)^2+zx(z+x)^2$$. You can rewrite the left hand side to obtain $$2(x^{4}+z^{4}) + 2(x^{4}+y^{4}) + 2(z^{4}+y^{4})\geq xy(x+y)^2+yz(y+z)^2+zx(z+x)^2$$. Let's take for example $$2(x^{4}+z^{4}) \geq xz(x+z)^2$$. If one of the terms $$x,z$$ is equal to $$0$$, then the inequality trivially holds. So if $$x,z\neq 0$$ we can write $$z=kx$$ to get the inequality $$2(1+k^{4})\geq k(k+1)^2$$ which seems to be true. Can you prove this one?

Hints towards an AM-GM solution.

1. Show that $$2x^4 + 2y^4 \geq xy(x+y)^2$$ by AM-GM.
2. Hence the first inequality is true.
3. Show that $$xy(x+y)^2 \geq 4x^2y^2$$ by AM-GM.
4. Show that $$x^2y^2 + y^2 z^2 \geq 2xy^2z$$ by AM-GM.
5. Hence, the second inequality is true.

Proof of the first part of the inequality:

Lemma: $$\mathbf {x^3+y^3 \ge xy(x+y)}$$

Using the lemma above: $$4(x^4+y^4+z^4) \ge \sum_{cyc} {\left(x^3+y^3\right)\left(x+y\right)}$$ $$4\left(\sum_{cyc}{x^4}\right) \ge 2\left(\sum_{cyc} x^4\right) + 2\left(\sum_{cyc} {x^3 \cdot y}\right)$$ $$2\left(\sum_{cyc} x^4\right) \ge 2\left(\sum_{cyc} {x^3 \cdot y}\right)$$ $$\sum_{cyc} x^4 \ge \sum_{cyc} {x^3 \cdot y}$$ This inequality is true according to the Rearrangement inequality.

Proof of the lemma:

$$\mathbf {x^3+y^3 \ge xy(x+y)}$$

Proof of the second part of the inequality:

From $$A.M \ge G.M \implies x+y \ge 2\sqrt{xy}$$ : $$4\cdot \sum_{cyc}{(xy)^2} \ge 4xyz(x+y+z)$$ $$\sum_{cyc}{(xy)^2} \ge xyz(x+y+z)$$ $$\sum_{cyc}{(xy)^2} \cdot (1+1+1) \ge 3\cdot xyz(x+y+z)$$ From Cauchy–Schwarz inequality , we have: $$(xy+yz+zx)^2 \ge 3yxz(x+y+z)$$ |$$xy=a$$|$$yz=b$$|$$zx=c$$|: $$(a+b+c)^2 \ge 3(ab+bc+ca)$$ $$a^2+b^2+c^2 \ge ab+bc+ca$$ Proof: $$a^2+b^2 \ge 2ab$$ $$b^2+c^2 \ge 2bc$$ $$a^2+b^2 \ge 2ab$$ Summing up these inequalities: $$a^2+b^2+c^2 \ge ab+bc+ca$$