Using $xy+yz+zx=1$, it suffices to prove \[ 2(xy+yz+zx)^3 + 9x^2y^2z^2 \ge 7xyz(x+y+z)(xy+yz+zx). \]Now, let $[m,n,k]$ be a shorthand for sums involving all terms of form $\alpha^m \beta^n \gamma^k$, where $(\alpha,\beta,\gamma)$ is a permutation of $(a,b,c)$. Expanding the LHS and doing some algebra, it boils down proving \[ 2[3,3,0]\ge [3,2,1], \]which follows immediately from Muirhead (alternative way is to finish with a judicious AM-GM).

Denote: $yz=a, zx=b, xy=c$. So, $a+b+c=1$. We need to prove that $2+9abc \geq 7(ab+bc+ca)$ or $2(a+b+c)^3+9abc\geq 7(ab+bc+ca)(a+b+c)$. This equivalent to $2a^3+2b^3+2c^3 \geq a^2b+ab^2+b^2c+bc^2+c^2a+ca^2$. It is very easy by AM-GM

We first clear the fractions and homogenize the inequality as \[2(xy+yz+zx)^3 + 9(xyz)^2 \ge 7(xyz)(x+y+z)(xy+yz+zx).\]Now, let $\tfrac{1}{x} + \tfrac{1}{y} + \tfrac{1}{z} = 1$, or $xy+yz+zx = xyz$, and simplify the above inequality to get \[2(xy+yz+zx) + 9 \ge 7(x+y+z).\]Then, homogenize again to get \[2(xy+yz+zx)\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)^2 + 9 \ge 7(x+y+z)\left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) .\]Expanding, this is equivalent to \[2\left( 3 \sum_{sym} \frac{x}{y} + \sum_{cyc} \frac{xy}{z^2} + 6 \right) + 9 \ge 7\left( 3+\sum_{sym} \frac{x}{y}\right),\]or \[\sum_{sym} \frac{xy}{z^2} \ge \sum_{sym} \frac{x}{y}.\]This is obvious by Muirhead, so we're done. $\blacksquare$

Orestis_Lignos wrote: For positive real numbers $x,y,z$ with $xy+yz+zx=1$, prove that $$\frac{2}{xyz}+9xyz \geq 7(x+y+z)$$ Let $ x,y,z>0 $ and $ xy+yz+zx=1. $ Prove that $$\frac{1}{xyz}+9xyz \geq 4(x+y+z)$$https://artofproblemsolving.com/community/c6h1343393p7304258 https://artofproblemsolving.com/community/c4h1400590p7832806

Can be solved in tons of ways since it is classical and homogenizable, here is another one: For convenience let $a=xy$, $b=yz$, $c=zx$, so $a+b+c=1$. We have $ab = xy^2z$, $bc=xyz^2$, $ac=x^2yz$, $abc = x^2y^2z^2$, hence we wish to prove $9x^2y^2z^2 + 2 \geq 7(x^2yz + xy^2z+xyz^2)$, i.e. $9abc + 2 \geq 7(ab+bc+ca)$. Let $s=a+b$, $p=ab$. Then $c=1-s$ and we want $9p(1-s) + 2 \geq 7p + 7s(1-s)$, i.e. $p(2-9s) \geq 7s - 7s^2 - 2$. At least one of the sums $a+b$, $b+c$, $c+a$ is at least $\frac{2}{3}$ (else the sum $a+b+c$ would be less than $1$), and by symmetry we may assume this holds for $a+b$, $s \geq \frac{2}{3} > \frac{2}{9}$. Hence we now wish to prove $\displaystyle p \leq \frac{7s^2 - 7s + 2}{9s-2}$. By the inequality $\displaystyle p \leq \frac{s^2}{4}$ (equivalent to $(x-y)^2 \geq 0$) it now suffices to have $\displaystyle \frac{s^2}{4} \leq \frac{7s^2 - 7s + 2}{9s-2}$. This is equivalent to $9s^3 - 30s^2 + 28s - 8 \leq 0$, i.e. to $(s-2)(3s-2)^2 \leq 0$, which holds, as $s < a+b+c = 1 < 2$. Equality holds if and only if $a+b = s = \frac{2}{3}$ and $a=b$, i.e. $a=b=c=\frac{1}{3}$, corresponding to $x=y=z=\frac{1}{\sqrt{3}}$.

My first post $$2 + 9(xyz)^2 \ge 7(xyz)(x + y + z)$$$$xy = a,\quad yz = b,\quad zx = c$$$$a+b+c = p = 1$$$$ab + bc + ca = q$$$$abc = r$$Then, we should prove $2 + 9r \ge 7q $ ; By Schur we know that $1 + 9r = p^3 + 9r \ge 4q$ ; So, we need prove that $p^2 = 1 \ge 3q$ this is also true. Done. Equality holds when:$$a=b=c=\frac{1}{3}$$$$xy=yz=zx=\frac{1}{3}$$$$x=y=z=\frac{1}{\sqrt{3}}$$

This was also on the Romania JTST and here is the solution I gave during the exam. It is sufficent to prove that $2 + 9xyz^{2} \ge 7(x+y+z)xyz$. Denote $a=xy$, $b=yz$, $c=zx$. The condition becomes $a+b+c = 1$ and the inequality to prove is $2+9abc \ge 7(ab+bc+ca)$. We (of course) homogenize and get: $$2(a+b+c)^{3} + 9abc \ge 7(ab+bc+ca)(a+b+c)$$ After expanding, this reduces to: $$2\sum_{cyc}a^{3} \ge \sum_{sym}a^{2}b$$which is true by Muirhead. Equality holds for $x=y=z=\frac{1}{\sqrt{3}}$

I apologize in advance. We use Lagrange Multipliers. Observe that $$f(x,y,z)=\frac{2}{xyz}+9xyz-7(x+y+z)$$has a global minimum for the given constraint due to compactness. For $g(x,y,z)=xy+yz+zx-1$ define the Lagrangian $$L(x,y,z,\lambda)=f(x,y,z)-\lambda g(x,y,z).$$Partially differentiating w.r.t. $x$ and equating to $0$ we get $$\frac{\partial L}{\partial x}=-\frac{2}{x^2yz}+9yz-7+\lambda(y+z)=0\Rightarrow 9x^2y^2z^2+\lambda(x^2yz)(y+z)-7x^2yz=2.$$Doing the same for other variables would yield the cyclic variations of this equation. Therefore we get, $$9x^2y^2z^2+\lambda(x^2yz)(y+z)-7x^2yz=9x^2y^2z^2+\lambda(xy^2z)(z+x)-7xy^2z=9x^2y^2z^2+\lambda(xyz^2)(x+y)-7xyz^2=2$$$$\Rightarrow (x-y)(\lambda z-7)=0, (y-z)(\lambda x-7)=0, (z-x)(\lambda y-7)=0$$if we pairwisely consider the equations. All of them must hold in order to achieve the minimum. This implies that for $k\in\mathbb{R}$ the only solution is $(x,y,z)=(k,k,k).$ Furthermore, taking the partial derivative with respect to $\lambda$ gives us that $$\frac{\partial L}{\partial \lambda}=xy+yz+zx-1=0\Rightarrow 3k^2=1\Rightarrow k=\frac{1}{\sqrt{3}}.$$At last we obtain that $$f(x,y,z)\geq f\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)=2\cdot 3\sqrt{3}+\sqrt{3}-7\cdot \sqrt{3}=0$$which proves the statement. $\blacksquare$