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Let $x,y,z>0$ Prove that:$$(x+y)(y+z)(z+x)\ge8(x+y+z)\sqrt[3]{x^2y^2z^2}$$

Again, I think of Schur, and the inequality is reversed again. By Schur, $$(x+y)(y+z)(z+x)\ge 8xyz$$ and we need to prove $$xyz\ge(x+y+z)\sqrt[3]{x^2y^2z^2}$$ But in fact, $$(x+y+z)\sqrt[3]{x^2y^2z^2}\ge xyz$$ I know my problem is if $a\ge b$,$a\ge c$, it isn't mean that $b\ge c$, please help me with this question and can you give me some experience to get out of this wrong way of thinking so that there can be many new directions?

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2 Answers2

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Proposition. $(x+y)(y+z)(z+x)\geq \frac 83(x+y+z)\sqrt[3]{x^2y^2z^2}$

To prove the Proposition, one uses the following lemma hinted by Paresseux Nguyen.

Lemma. $(x+y)(y+z)(z+x)\geq \frac 89(x+y+z)(xy+yz+zx)$

Proof. One uses the elementary symmetric polynomials: $$s_1=x+y+z,s_2=xy+yz+zx,s_3=xyz.$$ Then the statement in the Lemma is equivalent to $$9(s_1-x)(s_1-y)(s_1-z)\geq 8s_1s_2$$ $$\Leftrightarrow 9(s_1^3-s_1s_1^2+s_2s_1-s_3)\geq 8s_1s_2$$ $$\Leftrightarrow s_2s_1\geq 9s_3$$ $$\Leftrightarrow (xy+yz+zx)(x+y+z)\geq 9xyz,$$ which is true if one applies AM-GM twice. $\Box$

Proof of the Proposition. From the Lemma, it suffices to note that $$xy+yz+zx\geq 3\sqrt[3]{x^2y^2z^2},$$ by AM-GM. $\Box$

Pythagoras
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I think, it should be $$(x+y)(x+z)(y+z)\geq\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}.$$

Let $x+y+z=3u$, $xy+xz+yz=3v^2,$ where $v>0$ and $xyz=w^3$.

Thus, $u\geq v\geq w$ and we need to prove that: $$9uv^2-w^3\geq8uw^2$$ or $$9uv^2\geq8uw^2+w^3,$$ which is obvious because $uv^2\geq uw^2$ and $uv^2\geq w^3$.

Michael Rozenberg
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