Find the smallest real $ K$ such that for each $ x,y,z\in\mathbb R^{ + }$: \[ x\sqrt y + y\sqrt z + z\sqrt x\leq K\sqrt {(x + y)(y + z)(z + x)} \]
Problem
Source: Iranian National Olympiad (3rd Round) 2008
Tags: function, inequalities, inequalities proposed
31.08.2008 03:26
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.
31.08.2008 12:39
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
01.09.2008 08:56
Or you could do simple AM-GM like I did... lol.
16.09.2008 14:36
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$.
16.09.2008 19:17
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.
29.07.2020 21:16
29.07.2020 22:41
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
29.07.2020 23:10
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.
29.07.2020 23:29
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.)
29.07.2020 23:38
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.
01.05.2023 15:19
Using Cauchy Schwarz Inequality, $$ \begin{aligned} x\sqrt y + y\sqrt z + z\sqrt x &\leqslant\sqrt{(xy+yz+zx)(x+y+z)}\\ &=\sqrt{(x+y)(y+z)(z+x)+xyz}\\ &\leqslant\sqrt{\frac98(x+y)(y+z)(z+x)}\\ &=\boxed{\frac{3\sqrt 2}4}\sqrt {(x + y)(y + z)(z + x)}.\blacksquare \end{aligned}$$
01.05.2023 20:20
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 $9(x+y)(y+z)(z+x)\ge 8(x+y+z)(xy+yz+zx)$.