We have \[ LHS \ge \frac{\left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)^2}{x+y+z+3} \]Therefore, we need to prove that \[ \left( \frac{xy + yz + zx}{3(x+y+z)} \right)^2 \ge \frac{3(x+y+z + 3)}{4(x+y+z)} \]which is equivalent to \[ xy + yz + zx \ge \sqrt{\frac{27(x+y+z)(x+y+z+3)}{4}}\]But we know that $(xy + yz + zx)^2 \ge 3xyz(x+y+z) \ge (3(x+y+z))^2 $ and denoting $x+y+z = a$. It is equivalent to proving \[ 36a^2 \ge 27(a^2 + 3a) \Rightarrow a \ge 9 \]But this is clearly true since $xyz = 3(x+y+z) \ge 9 \sqrt[3]{xyz} \Rightarrow xyz \ge 27$, which forces $x + y + z = \frac{xyz}{3} \ge 9$.

Notice that the inequality is equivalent to: $\sum_{cyc}\frac{4yz}{xy+x}\ge9$ by Cauchy-Swartz we have that: $\sum_{cyc}\frac{4yz}{xy+x}\ge\frac{16(\sum_{cyc}xy)^2}{4(\sum_{cyc}yz(xy+x))}$ Furthermore the inequality boils down to: $\frac{4(\sum_{cyc}xy)^2}{3xyz+xyz\sum_{cyc}x}\ge9$ $\Longrightarrow \frac{4(\sum_{cyc}xy)^2)}{9xyz+x^2y^2z^2}\ge3$ Thus we need to prove: $4(\sum_{cyc}xy)^2\ge 18xyz+3x^2y^2z^2\Longrightarrow 4\sum_{cyc}x^2y^2+8xyz\sum_{cyc}x\ge 18xyz+3x^2y^2z^2$ Which is clearly true since: $\sum_{cyc}x\ge9$ and $xyz\ge 27$ from our conditions. And we are done!

x+y+z≥（xy+yz+zx）/3，it is wrong！

ylmath123 wrote: x+y+z≥（xy+yz+zx）/3，it is wrong！ You are right I missed some signs.

IndoMathXdZ wrote: We have \[ LHS \ge \frac{\left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)^2}{x+y+z+3} \]Therefore, we need to prove that \[ \left( \frac{xy + yz + zx}{3(x+y+z)} \right)^2 \ge \frac{3(x+y+z + 3)}{4(x+y+z)} \]which is equivalent to \[ xy + yz + zx \ge \sqrt{\frac{27(x+y+z)(x+y+z+3)}{4}}\]But we know that $(xy + yz + zx)^2 \ge 3xyz(x+y+z) \ge (3(x+y+z))^2 $ and denoting $x+y+z = a$. It is equivalent to proving \[ 36a^2 \ge 27(a^2 + 3a) \Rightarrow a \ge 9 \]But this is clearly true since $xyz = 3(x+y+z) \ge 9 \sqrt[3]{xyz} \Rightarrow xyz \ge 27$, which forces $x + y + z = \frac{xyz}{3} \ge 9$. ohh u missed a clear chance btw very good sol xy+yz+zx=xyz(1/x+1/y+1/z)>=xyz and 1/x+1/y+1/z>=1 we have (1/x+1/y+1/z)^2*4(x+y+z)>=3*(x+y+z+3) then x+y+z>=9 which is obviously true

Multiplying both sides by $(x+y+z+3)$, by Cauchy-Schwarz it suffices to show that $$\frac{3(x+y+z+3)}{4(x+y+z)} \leq \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\left(\frac{xy+yz+zx}{xyz}\right)^2=\left(\frac{xy+yz+zx}{3(x+y+z)}\right)^2 \iff 4(xy+yz+zx)^2\geq 27(x+y+z)(x+y+z+3)=9xyz(x+y+z+3).$$We expand the RHS as $4(x^2y^2+y^2z^2+z^2x^2)+8xyz(x+y+z)$, so the inequality reduces to $4(x^2y^2+y^2z^2+z^2x^2) \geq xyz(x+y+z)+27xyz$ Since $x^2y^2+y^2z^2+z^2x^2+ \geq xyz(x+y+z)$ by Muirhead, it suffices to show that $$x^2y^2+y^2z^2+z^2x^2\geq 9xyz=9xyz\sqrt{\frac{xyz}{3(x+y+z)}} \iff (x^2y^2+y^2z^2+z^2x^2)\sqrt{3(x+y+z)} \geq 9(xyz)^{3/2},$$which is true by AM-GM on the separate terms of the product. $\blacksquare$ Remark: It is also possible to use AM-GM first to reduce to $xyz \geq 27$, which is true since $xyz=3(x+y+z) \geq 9\sqrt[3]{xyz}$.

Hopefully not a fakesolve, otherwise someone please point it out. Sub $a=\frac{1}{x}$, $b=\frac{1}{y}$, $c=\frac{1}{z}$. The condition reduces down to $\frac{1}{abc}=3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$, or $ab+bc+ca=\frac{1}{3}$, whilst the inequality itself reduces down to \begin{align*} \sum_{\textrm{cyc}}\frac{a^2b}{b+1}&\ge\frac{3}{4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\\ \sum_{\textrm{cyc}}\frac{a^2b}{b+1}&\ge\frac{3}{4\frac{ab+bc+ca}{abc}}\\ \sum_{\textrm{cyc}}\frac{a^2b}{b+1}&\ge\frac{9abc}{4}\\ \sum_{\textrm{cyc}}\frac{a}{c(b+1)}&\ge\frac{9}{4} \end{align*}However by Cauchy Schwarz: \begin{align*} \sum_{\textrm{cyc}}\frac{a}{c(b+1)}&=\sum_{\textrm{cyc}}\frac{a^2}{ac(b+1)}\\ &\ge\frac{(a+b+c)^2}{3abc+(ab+bc+ca)}\\ &=\frac{(a+b+c)^2}{3abc+\frac{1}{3}} \end{align*}However $(a+b+c)^2\ge3(ab+bc+ca)=1$, whilst $\frac{1}{9}=\frac{ab+bc+ca}{3}\ge\sqrt[3]{(abc)^2}$, so $abc\le\frac{1}{27}$. Hence: \begin{align*} \sum_{\textrm{cyc}}\frac{a^2}{bc(b+1)}&\ge\frac{1}{3\cdot\frac{1}{27}+\frac{1}{3}}=\frac{1}{\frac{1}{9}+\frac{1}{3}}=\frac{1}{\frac{4}{9}}=\frac{9}{4} \end{align*}as desired.

If we look at the inequality by zoom out; (So simple solution) Just apply $AM-GM$. If we Make it more simple,It will be; $64*x*y*z \ge 27*(x+1)*(y+1)*(z+1)$ After simplifizing; $28*(x+y+z) \ge 9(xy+yz+zx) +9$ (2) Then, for the given thing, we have to make it more useful. $x*y*z=3*(x+y+z)$ divide by xyz(simplifize too); $\frac{1}{3}=\frac{1}{x*y}+\frac{1}{y*z}+\frac{1}{z*x}$(1) Apply Titu to (1), We'll find; $27 \ge x*y+y*z+z*x$ (3) From there also we have; $x+y+z \ge 9$ (4) For (2) use (3) and (4) , we proved. It holds for $(3,3,3)$