Prove $\sqrt[4]{\frac{xy}{z}}+\sqrt[4]{\frac{zy}{x}}+\sqrt[4]{\frac{xz}{y}}\ge 3.$

Question

Problem. Let $x,y,z>0$ and $x+y+z=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}.$ Prove that$$\sqrt[4]{\frac{xy}{z}}+\sqrt[4]{\frac{zy}{x}}+\sqrt[4]{\frac{xz}{y}}\ge 3.$$ Equality holds iff $x=y=z=1.$

I used the substitution $xy:=a,\ yz:=b,\ zx:=c$ then $a,b,c>0:a+b+c=ab+bc+ca.$

The starting inequality becomes$$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 3\sqrt[8]{abc},$$or $$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 3\sqrt[8]{abc\cdot\frac{ab+bc+ca}{a+b+c}}.$$

Could you please give me a hint to continue? Thank you for your help.

PS. By my attempt, I've tried to finish my idea.

We'll prove $$(a+b+c)^8(a^2+b^2+c^2)\ge 3^8a^2b^2c^2(a^2b^2+b^2c^2+c^2a^2).\label{1}\tag{1}$$ Since $$(a^2+b^2+c^2)^2\ge 3(a^2b^2+b^2c^2+c^2a^2),$$it's enough to prove$$(a+b+c)^8\ge 3^7a^2b^2c^2(a^2+b^2+c^2).\label{2}\tag{2}$$ By AM-GM inequality $$ \begin{align} (a+b+c)^6\ge 27(a^2+b^2+c^2)(ab+bc+ca)^2 &\ge 81abc(a+b+c)(a^2+b^2+c^2)\\ \implies (a+b+c)^5&\ge 81abc(a^2+b^2+c^2).\label{3}\tag{3} \end{align} $$ Also,$$(a+b+c)^3\ge 27abc.\label{4}\tag{4}$$ From \eqref{3} and \eqref{4}, we obtain the desired inequality \eqref{2}.

Answer 1

uvw helps.

Indeed, by using your work we need to prove $$\left(\frac{a+b+c}{3}\right)^8(a^2+b^2+c^2)\geq(a^2b^2+a^2b^2+b^2c^2)a^2b^2c^2,$$ which in the uvw's notation gives $f(u)\geq0,$ where $$f(u)=u^8(3u^2-2v^2)-w^6(3v^4-2uw^3).$$ But by Maclaurin $$f'(u)=30u^9-16u^7v^2+2w^9\geq0,$$ which says that $f$ increases and since the last inequality is homogeneous, by uvw it's enough to prove the last inequality for $b=c=1$, which gives $$\left(\frac{a+2}{3}\right)^8(a^2+2)\geq(2a^2+1)a^2.$$ Can you finish it now?

Answer 2

PS. By my attempt, I've tried to finish my idea.

We'll prove $$(a+b+c)^8(a^2+b^2+c^2)\ge 3^8a^2b^2c^2(a^2b^2+b^2c^2+c^2a^2).\label{1}\tag{1}$$ Since $$(a^2+b^2+c^2)^2\ge 3(a^2b^2+b^2c^2+c^2a^2),$$it's enough to prove$$(a+b+c)^8\ge 3^7a^2b^2c^2(a^2+b^2+c^2).\label{2}\tag{2}$$ By AM-GM inequality $$ \begin{align} (a+b+c)^6\ge 27(a^2+b^2+c^2)(ab+bc+ca)^2 &\ge 81abc(a+b+c)(a^2+b^2+c^2)\\ \implies (a+b+c)^5&\ge 81abc(a^2+b^2+c^2).\label{3}\tag{3} \end{align} $$ Also,$$(a+b+c)^3\ge 27abc.\label{4}\tag{4}$$ From \eqref{3} and \eqref{4}, we obtain the desired inequality \eqref{2}.

Answer 3

You are almost there. apply the am-gm inequality to get the answer-- take $\sqrt a, \sqrt b, \sqrt c$ as $ a_1, a_2, a_3 $ respectively. Therefore, we know that $(a_1+a_2+a_3)/3 \geq (a_1a_2a_3)^{1/3}$. But in your second-to-last step, our sum of $a_i$ should be $\geq$ the $1/4$th power of their product. Hence proved.

Answer 4

The pqr method works well.

Proceeding along OP's lines, it suffices to prove that $$(a+b+c)^8(a^2+b^2+c^2)\ge 3^8a^2b^2c^2(a^2b^2+b^2c^2+c^2a^2). \tag{1}$$

Let $p := a + b + c, q := ab + bc + ca, r := abc$. We have $p^2 \ge 3q$ and $p^3 \ge 27r$ (by AM-GM).

(1) is written as $$p^8 (p^2 - 2q) \ge 3^8 r^2 (q^2 - 2pr).$$

Using $p^2 \ge 3q$, it suffices to prove that $$p^8 (p^2 - 2\cdot p^2/3) \ge 3^8 r^2 (p^4/9 - 2pr),$$ or $$\frac13 p^{10} \ge 3^8 p r^2 (p^3/9 - 2r).$$

By AM-GM, we have $$3^8 p r^2 (p^3/9 - 2r) = 3^8 p \cdot r \cdot r \cdot (p^3/9 - 2r) \le 3^8p\left(\frac{r + r + p^3/9 - 2r}{3}\right)^3 = \frac13 p^{10}.$$

We are done.