$y(x+1) + 2z \overset{AM-GM}\ge 2\sqrt{2yz(x+1)} \ge 2\sqrt{4xyz}$, because $2(x+1) \ge 4x$ as $1 \ge x$.

Solution from PuzzlingCane We can rewrite $xy+y+2z$ as $(x-1)y+2y+2z$. Since $x-1$ is not positive and $y$ is positive, $(x-1)y$ is not positive, which means that $(x-1)y+2y+2z\ge 2y+2z$. By AGM, $2y+2z\ge 2\sqrt{2y*2z}=4\sqrt{yz}$. Since $0<x\leq 1$, $0<\sqrt{x}\leq 1$ Therefore, $4\sqrt{yz}\ge 4\sqrt{yz}*\sqrt{x}=4\sqrt{xyz}$. Therefore, we are done.

BarisKoyuncu wrote: Solution from PuzzlingCane We can rewrite $xy+y+2z$ as $(x-1)y+2y+2z$. Since $x-1$ is not positive and $y$ is positive, $(x-1)y$ is not positive, which means that $(x-1)y+2y+2z\ge 2y+2z$. By AGM, $2y+2z\ge 2\sqrt{2y*2z}=4\sqrt{yz}$. Since $0<x\leq 1$, $0<\sqrt{x}\leq 1$ Therefore, $4\sqrt{yz}\ge 4\sqrt{yz}*\sqrt{x}=4\sqrt{xyz}$. Therefore, we are done. If $(x-1)y \le 0$, then $(x-1)y + 2y + 2z \le 0 + 2y + 2z \le 2y+2z$ and not the other way around

BarisKoyuncu wrote: Solution from PuzzlingCane We can rewrite $xy+y+2z$ as $(x-1)y+2y+2z$. Since $x-1$ is not positive and $y$ is positive, $(x-1)y$ is not positive, which means that $(x-1)y+2y+2z\ge 2y+2z$. By AGM, $2y+2z\ge 2\sqrt{2y*2z}=4\sqrt{yz}$. Since $0<x\leq 1$, $0<\sqrt{x}\leq 1$ Therefore, $4\sqrt{yz}\ge 4\sqrt{yz}*\sqrt{x}=4\sqrt{xyz}$. Therefore, we are done. Amazing solution

Fairly certain this works given how similar it is to the other solutions: $xy+y+2z >= (1+x)(y+z)$ and by AM-GM $1+x>=2\sqrt x$ and $y+z >= 2\sqrt{yz}$ so $$xy+y+2z>=(1+x)(y+z) >= 4\sqrt{xyz}$$

Let $x, y, z$ are positive reals such that $x \leq 1$. Prove that $$xy+y+kz \geq 2 \sqrt{2kxyz}$$Where $k>0.$

Let $x, y, z$ are positive reals such that $x \leq 1$. Prove that $$xy+k_1y+k_2z \geq 2 \sqrt{k_2(k_1+1)xyz}$$Where $k_1,k_2>0.$

My solution during the contest. We can easily say that $xy+y+2z=xy+y+z+z$ By AM-GM We get $xy+y+z+z \geq 4 \sqrt{yz} \sqrt[4]{x}$ We know that $\sqrt[4]{x} \geq \sqrt{x}$ So $xy+y+2z=xy+y+z+z \geq 4 \sqrt{yz} \sqrt[4]{x} \geq 4 \sqrt{xyz}$

hhhhmmm isn't it just $xy + y + 2z \geq xy + xy + 2z = 2xy + 2z \geq 4\sqrt {xyz}$ since $x \leq 1 \Leftrightarrow xy \leq y$ ?

My solution during the contest: If $x \leq 1$ , we can see that $y \geq xy$. Hence, $xy+y+2z \geq 2xy+2z$. By AM-GM: $$xy+y+2z \geq 2xy+2z \geq 4\sqrt{xyz}$$

sqing wrote: Let $x, y, z$ are positive reals such that $x \leq 1$. Prove that $$xy+k_1y+k_2z \geq 2 \sqrt{k_2(k_1+1)xyz}$$Where $k_1,k_2>0.$

My solution $xy+y+2z\geq4\sqrt{xyz}$ $xy+y+2z\geq2(xy+z)$ $y\geq xy$ $1\geq x$ And we are done

youthdoo wrote:

I think this (also in this topic): By AM-GM, $y(x+1) + 2z \ge 2\sqrt{y(x+1)\cdot 2z} \ge 2\sqrt{2yz\cdot 2x} = 2\sqrt{4xyz}$ where we use $x+1 \ge 2x$.

$\color{blue} \boxed{\textbf{SOLUTION}}$ By $\color{red} \textbf{AM-GM Inequality,}$ $$xy+z \geq 2 \sqrt{xyz}$$$$y+z \geq 2 \sqrt{yz} \geq 2 \sqrt{xyz}, [x \leq 1]$$Adding these gives $$xy+y+2z \geq 4 \sqrt{xyz}$$$\blacksquare$

$xy+y+2z=xy+y+z+z=xy+z+y+z$ By $AM-GM$ $xy+z\geq 2 \sqrt{xyz}$ $y+z\geq 2 \sqrt{yz}= 2 \sqrt{yz*1}\geq 2 \sqrt{yz*x}= 2 \sqrt{xyz}$ $xy+z+y+z\geq 2 \sqrt{xyz}+2 \sqrt{xyz}$ $xy+y+2z\geq4\sqrt{xyz}$

Notice AM-GM yields \[ xy + y + 2z = y(x+1) + 2z \ge 2 \sqrt{2yz(x+1)} \ge 2 \sqrt{2yz(2x)} = 4 \sqrt{xyz} \]where the second inequality follows from $x \le 1$. $\blacksquare$ Remark: We are motivated to factor $xy + y$ as $y(x+1)$ in order to ensure the powers of $x, y, z$ are equal when applying AM-GM.