Very elegant problem.

Let $x,y,z$ are positive real numbers such that $xyz+x+y+z=6$ and $xyz+2xy+yz+zx+z=10.$ Prove that $$(xy+1)(yz+1)(zx+1)\leq 12(4 - \sqrt{6}) .$$

Original Problem wrote: Let $x,y,z$ are positive real numbers such that $xyz+x+y+z=6$ and $xyz+2xy+yz+zx+z=10.$ Find the maximum value of \[ (xy + 1)(yz + 1)(zx + 1)\] Annoying problem! The key of this problem is really tricky. Notice that \[ xyz + 2xy + yz + zx + z + 2 = 2(xyz + x + y + z) \]This simplifies to $(x-1)(y-1)(z-2) = 0$. If $x = 1$, then we have the condition $yz + y + z = 5$ and we would like to maximize $(y + 1)(z + 1)(yz + 1)$ which is \[ (y+1)(z+1)(yz+1) = 6(yz+1) \le 12(4 - \sqrt{6}) \]The last part come from the fact that $2\sqrt{yz} + yz \le yz + y + z = 5$ gives us $yz \le (\sqrt{6} - 1)^2 = 7 - 2 \sqrt{6}$. Equality holds when $(x,y,z) = (1, \sqrt{6} - 1, \sqrt{6} - 1)$. A similar results hold for $y = 1$ since the expression given is symmetric wrt to $x$ and $y$. (NB : because swapping $x$ and $y$ in any of the equations won't change the result.) If $z = 2$, we then have $2xy + x + y = 4$ and we would like to maximize \[ (xy + 1)(2y + 1)(2x + 1) = (xy + 1)(2(2xy + x + y) + 1) = 9(xy + 1) \]But, we have $2\sqrt{xy} + 2xy \le 2xy + x + y = 4$, which gives us $xy \le 1$. Then this gives us the maximum value as $18$ which holds for $(x,y,z) = (2,1,1)$. We have exhaust all cases and therefore the maximum value of the given expression is $\boxed{12 (4 - \sqrt{6})}$

or $\left\{\begin{array}{l} 2xy+(z-1)(x+y)=4\\ zxy+(x+y)=6-z \end{array}\right.\Longleftrightarrow \left\{\begin{array}{l} xy=\frac{5-z}{z+1}\\ x+y=\frac{6}{z+1} \mbox{or} z=2 \end{array}\right.$ so $(xy+1)(yz+1)(zx+1)=\frac{6}{z+1}\left(z^2xy+(x+y)z+1\right)=\frac{6(-z^2+6z+1)}{z+1}=6\left(8-(z+1)-\frac{6}{z+1}\right)\le 6\left(8-2\sqrt{6}\right).$

Let $a,b,c$ be real numbers such that $a+b+c=0$ and $a^4+b^4+c^4=18.$ Prove that$$a^5+b^5+c^5\leq 30.$$Let $ a,b,c>0, abc=1$ and $ab+bc+ca=5. $ Prove that$$\frac{17}{4} \leq a+b+c \leq 1+ \sqrt{32}.$$

Can you post P4,P5 here?Thanks

sqing wrote: Let $x,y,z$ are positive real numbers such that $xyz+x+y+z=6$ and $xyz+2xy+yz+zx+z=10.$ Prove that $$(xy+1)(yz+1)(zx+1)\leq 12(4 - \sqrt{6}) .$$ $$(x-1)(y-1)(z-2)=2(xyz+x+y+z-1)-(xyz+2xy+yz+zx+z)=0$$h h

sqing wrote:

Let $a,b,c$ be real numbers such that $a+b+c=0$ and $a^4+b^4+c^4=18.$ Prove that$$a^5+b^5+c^5\leq 30.$$Let $ a,b,c>0, abc=1$ and $ab+bc+ca=5. $ Prove that$$\frac{17}{4} \leq a+b+c \leq 1+ \sqrt{32}.$$ Hi, wonderful problems indeed. And happy to see you safe from Coronavirus @sqing sir. 预防冠状病毒

Let $a,b,c$ be real numbers such that $a^2+2b^2+3c^2=36. $ Prove that$$ab+bc+ca+a+20b+51c\leq 205.$$

Let $ a,b,c\ge 0, abc+a+b=4$ and $abc+ab+2c=7. $ Prove that $$(ab+1)(bc+1)(ca+1) \leq 18$$Let $ a,b,c\ge 0$ and $a+b+bc+ca=ab+c+abc=1. $ Prove that $$(ab+1)(bc+1)(ca+1) \leq 38-21\sqrt 3$$

sqing wrote: Let $a,b,c$ be real numbers such that $a^2+2b^2+3c^2=36. $ Prove that$$ab+bc+ca+a+20b+51c\leq 205.$$ $$205 - (ab+bc+ca+a+20b+51c) + 3(a^2+2b^2+3c^2 - 36) $$$$= \frac{1}{12}\, \left( 6\,a-b-c-1 \right) ^{2}+{\frac {1}{852}}\, \left( 71\,b- 7\,c-121 \right) ^{2}+{\frac {629}{71}}\, \left( c-3 \right) ^{2}.$$

Let $ a,b,c> 0, abc=1$ and $a+b+c=1+4\sqrt{2}. $ Prove that$$ab+bc+ca\geq 5$$$$(ab+1)(bc+1)(ca+1)\ge 8+4\sqrt 2$$Let $ a,b,c> 0, abc=1$ and $a+b+c=\frac{17}{4}. $ Prove that$$ab+bc+ca\leq 5$$$$(ab+1)(bc+1)(ca+1) \leq \frac{45}{4}$$

sqing wrote: Let $ a,b,c> 0, abc=1$ and $a+b+c=1+4\sqrt{2}. $ Prove that$$ab+bc+ca\geq 5$$$$(ab+1)(bc+1)(ca+1)\ge 8+4\sqrt 2$$