We want to find the smallest K. I claim $ K=3/(2\sqrt{2})$ $ 8(x\sqrt{y}+y\sqrt{z}+z\sqrt{x})^2 \le 9(x+y)(y+z)(z+x)$ $ 8x^2y+8y^2z+8z^2x+16xy\sqrt{yz}+16yz\sqrt{zx}+16xz\sqrt{xy} \le 9\sum_{sym}x^2y +18xyz$ $ 16xy\sqrt{yz}+16yz\sqrt{zx}+16xz\sqrt{xy} \le x^2y+y^2z+z^2x+9y^2x+9z^2y+9x^2z+18xyz$ By AM-GM $ z^2x+(9)y^2x+(6)xyz\ge 16\sqrt[16]{z^2x*y^{18}x^9*x^6y^6z^6}=16xy\sqrt{xz}$ Sum up cyclically. We can get equality when x=y=z=1, so we know that K cannot be any smaller.

rofler wrote: We want to find the smallest K. I claim $ K = 3/(2\sqrt {2})$ $ 8(x\sqrt {y} + y\sqrt {z} + z\sqrt {x})^2 \le 9(x + y)(y + z)(z + x)$ $ 8x^2y + 8y^2z + 8z^2x + 16xy\sqrt {yz} + 16yz\sqrt {zx} + 16xz\sqrt {xy} \le 9\sum_{sym}x^2y + 18xyz$ $ 16xy\sqrt {yz} + 16yz\sqrt {zx} + 16xz\sqrt {xy} \le x^2y + y^2z + z^2x + 9y^2x + 9z^2y + 9x^2z + 18xyz$ By AM-GM $ z^2x + (9)y^2x + (6)xyz\ge 16\sqrt [16]{z^2x*y^{18}x^9*x^6y^6z^6} = 16xy\sqrt {xz}$ Sum up cyclically. We can get equality when x=y=z=1, so we know that K cannot be any smaller. There are many way too kill it Use pqr Or Use ABC theorem :Fix $ x+y+z,\sum xy$ then it is a crease function with $ xyz$ so we only need check with $ x=y$after some small caculator we will find $ K$ followed above

Or you could do simple AM-GM like I did... lol.

My solution, also with AM-GM: We want to find the smallest K such that $ (x\sqrt {y} + y\sqrt {z} + z\sqrt {x})^2 \leq K^2(x + y)(y + z)(z + x)$ But $ (x\sqrt {y} + y\sqrt {z} + z\sqrt {x})^2 = \sum_{cyc}x^2y + 2(\sum_{cyc} xy\sqrt {yz}) \leq \sum_{cyc}x^2y + 2(\sum_{cyc} \frac {xyz + xy^2}{2} ) = (x + y)(y + z)(z + x) + xyz \leq (x + y)(y + z)(z + x) + \frac {1}{8}(x + y)(y + z)(z + x) = \frac {9}{8}(x + y)(y + z)(z + x)$ Therefore, $ K^2\geq \frac {9}{8} \rightarrow K \geq \frac {3}{2\sqrt {2}}$ with equaltiy if and only if $ x = y = z$.

Omid Hatami wrote: Find the smallest real $ K$ such that for each $ x,y,z\in\mathbb R^{ + }$: \[ x\sqrt y + y\sqrt z + z\sqrt x\leq K\sqrt {(x + y)(y + z)(z + x)} \] My solution using Cauchy-Schwarz: We have \begin{align*}LHS & = \sqrt x\sqrt {xy} + \sqrt y\sqrt {yz} + \sqrt z\sqrt {zx} \\ & \le\sqrt {\left(x + y + z\right)\left(xy + yz + zx\right)} \\ & \le\frac {3}{2\sqrt 2}\sqrt {(x + y)(y + z)(z + x)}\end{align*} where the last inequality follows from $ 8(x + y + z)(xy + yz + zx)\le 9(x + y)(y + z)(z + x)$, which is well known.

Allnames wrote: rofler wrote: We want to find the smallest K. I claim $ K = 3/(2\sqrt {2})$ $ 8(x\sqrt {y} + y\sqrt {z} + z\sqrt {x})^2 \le 9(x + y)(y + z)(z + x)$ $ 8x^2y + 8y^2z + 8z^2x + 16xy\sqrt {yz} + 16yz\sqrt {zx} + 16xz\sqrt {xy} \le 9\sum_{sym}x^2y + 18xyz$ $ 16xy\sqrt {yz} + 16yz\sqrt {zx} + 16xz\sqrt {xy} \le x^2y + y^2z + z^2x + 9y^2x + 9z^2y + 9x^2z + 18xyz$ By AM-GM $ z^2x + (9)y^2x + (6)xyz\ge 16\sqrt [16]{z^2x*y^{18}x^9*x^6y^6z^6} = 16xy\sqrt {xz}$ Sum up cyclically. We can get equality when x=y=z=1, so we know that K cannot be any smaller. There are many way too kill it Use pqr Or Use ABC theorem :Fix $ x+y+z,\sum xy$ then it is a crease function with $ xyz$ so we only need check with $ x=y$after some small caculator we will find $ K$ followed above I am new to inequalities and would like to know what is that treorem you called, the ABC theorem

I searched "ABC theorem" and didn't find anything about it. I got something about the ABC conjecture (which is an interesting one) but I know that's not what's used here since that remains unproven.

Search uvw method instead. You’ll find similar things. (But as someone mentioned in some other Iran forum, you cannot use UVW,ABC method in Iran.)

Oh, right. I've heard of the uvw method being mentioned before, and I felt like that was what was being mentioned, but I forgot which 3 letters they were.

Using Cauchy Schwarz Inequality, $$ \begin{aligned} x\sqrt y + y\sqrt z + z\sqrt x &\leqslant\sqrt{(xy+yz+zx)(x+y+z)}\\ &=\sqrt{(x+y)(y+z)(z+x)+xyz}\\ &\leqslant\sqrt{\frac98(x+y)(y+z)(z+x)}\\ &=\boxed{\frac{3\sqrt 2}4}\sqrt {(x + y)(y + z)(z + x)}.\blacksquare \end{aligned}$$

A simple argument. Taking $x=y=z$, we get $K\ge K^* = 3/(2\sqrt{2})$. We now show the inequality with $K^*$. Without loss of generality, let $x+y+z=1$. Using concavity of $x\mapsto \sqrt{x}$, we have $x\sqrt{y}+y\sqrt{z}+z\sqrt{z}\le \sqrt{xy+yz+zx} = \sqrt{(xy+yz+zx)(x+y+z)}$. From here, the result follows using the well-known $9(x+y)(y+z)(z+x)\ge 8(x+y+z)(xy+yz+zx)$.