INAMO 2021/4 wrote: Let $x,y$ and $z$ be positive reals such that $x + y + z = 3$. Prove that \[ 2 \sqrt{x + \sqrt{y}} + 2 \sqrt{y + \sqrt{z}} + 2 \sqrt{z + \sqrt{x}} \le \sqrt{8 + x - y} + \sqrt{8 + y - z} + \sqrt{8 + z - x} \] This gives me ISL 2020 A4 and A3 nightmares. Cyclic inequalities should be hard right? Apparently AM-GM and CS is enough: \begin{align*} 2 \sum_{cyc} \sqrt{x + \sqrt{y}} &\le 2 \sum_{cyc} \sqrt{x + \frac{y + 1}{2}} \\ &= \sum_{cyc} \sqrt{x + \frac{y + 1}{2}} + \sqrt{y + \frac{z + 1}{2}} \\ &\le \sum_{cyc} \sqrt{2 \left( x + \frac{y + 1}{2} + y + \frac{z + 1}{2} \right)} \\ &= \sum_{cyc} \sqrt{2(x + y) + (y + z + 2)} \\ &= \sum_{cyc} \sqrt{2(3 - z) + (y + z + 2)} = \sum_{cyc} \sqrt{8 + y - z} \end{align*}

GorgonMathDota wrote: Let $x,y$ and $z$ be positive reals such that $x + y + z = 3$. Prove that \[ 2 \sqrt{x + \sqrt{y}} + 2 \sqrt{y + \sqrt{z}} + 2 \sqrt{z + \sqrt{x}} \le \sqrt{8 + x - y} + \sqrt{8 + y - z} + \sqrt{8 + z - x} \]

Attachments:

Let $x,y$ and $z$ be positive reals such that $x + y + z = 3$. Prove that $$2 \sqrt{x + \sqrt{y}} + 2 \sqrt{y + \sqrt{z}} + 2 \sqrt{z + \sqrt{x}} \le \sqrt{8 + x - y} + \sqrt{8 + y - z} + \sqrt{8 + z - x}\leq 6\sqrt 2$$

My solution during contest is almost similar to @IndoMathXdZ. By AM-GM, we have \begin{align*} \sum_{\sigma}2\sqrt{x+\sqrt{y}} &\le \sum_{\sigma}2 \sqrt{x+\frac{y+1}{2}}\\ &=\sum_{\sigma}\sqrt{ 4x+2y+2}\\ &= \sum_{\sigma} \sqrt{8+2x-2z}. \end{align*}By AM-QM: \begin{align*} \sum_{\sigma} \sqrt{8+2x-2z} &= \frac{1}{2}\sum_{\sigma} \left (\sqrt{8+2x-2z} + \sqrt{8+2z-2y} \right ) \\ &\le \frac{1}{2}\sum_{\sigma}2\sqrt{\frac{8+2x-2z+8+2z-2y}{2}}\\ &\le \sum_{\sigma} \sqrt{8+x-y}. \end{align*}Therefore, \[2\sum_{\sigma} \sqrt{x+\sqrt{y}}\le \sum_{\sigma} \sqrt{8+2x-2z} \le \sum_{\sigma} \sqrt{8+x-y}\implies 2\sum_{\sigma} \sqrt{x+\sqrt{y}} \le \sum_{\sigma}\sqrt{8+x-y}.\]Equality occurs when $x=y=z=1$.

$\sqrt{y} \leq \frac{y+1}{2}$ is actually not needed. By QM-AM: \begin{align*} LHS = \sum_{cyc} \sqrt{x + \sqrt{y}} + \sqrt{y + \sqrt{z}} &\leq \sum_{cyc} \sqrt{2(x + y + \sqrt{y} + \sqrt{z})} \\ &= \sum_{cyc} \sqrt{6 - 2z + 2\sqrt{y} + 2\sqrt{z}} \\ &\leq \sum_{cyc} \sqrt{8 + y - z} \end{align*} The last inequality follows from $(\sqrt{z} - 1)^2 + (\sqrt{y} - 1)^2 \geq 0$. Edit @Below oh yes

Isn't the last line basically the same thing though?

notice $x-y = 3-2y-z$ $y-z = 3-2z-x$ $z-x = 3-2x-y$ $2\sqrt{x+\sqrt{y}} \leq^{AM-GM} 2\sqrt{x+\frac{y+1}{2}}$ $2\sqrt{y+\sqrt{z}} \leq^{AM-GM} 2\sqrt{y+\frac{z+1}{2}}$ $2\sqrt{z+\sqrt{x}} \leq^{AM-GM} 2\sqrt{z+\frac{x+1}{2}}$ pick each 2 perms of $(\sqrt{x+\frac{y+1}{2}},\sqrt{y+\frac{z+1}{2}},\sqrt{z+\frac{x+1}{2}})$ and use Cauchy $(\sqrt{x+\frac{y+1}{2}}+\sqrt{y+\frac{z+1}{2}})\leq (1+1)^{\frac{1}{2}}(x+\frac{3y}{2}+1+\frac{z}{2})^{\frac{1}{2}}=\sqrt{(2x+3y+2+z)}=\sqrt{x+2y+5} =\sqrt{8-y-z}$ analogous for the other 2 perms