三元不等式

大佬最帅

$设a，b，c> 0，求证：\sum \sqrt{\frac{a}{a+  2b+  c} } \le \frac{3}{2}$

hbghlyj

We want to show
\[\sum\sqrt{\dfrac{x}{x+y+2z}}\leq\dfrac{3}{2}\]We can use the Cauchy-Schwarz Inequality to get
\begin{aligned}
&\left(\sum \left(x(x+2y+z)\right)\left(\frac{1}{(x+2y+z)(x+y+2z)}\right)\right)^2\\
\leq&\left(\sum x^2+2xy+xz\right)\left(\sum\frac{1}{(x+2y+z)(x+y+2z)}\right)\\
=&\frac{4\left(\sum x^2+3\sum xy\right)\left(\sum x\right)}{(x+y+2z)(x+2y+z)(2x+y+z)}
\end{aligned}We thus want to show that
\begin{multline*}16\left(x^3 + 4 x^2 y + 4 x^2 z + 4 x y^2 + 9 x y z + 4 x z^2 + y^3 + 4 y^2 z + 4 y z^2 + z^3\right)\\\leq 9\left(2 x^3 + 7 x^2 y + 7 x^2 z + 7 x y^2 + 16 x y z + 7 x z^2 + 2 y^3 + 7 y^2 z + 7 y z^2 + 2 z^3\right)\end{multline*}This rearranges to give
\[2x^3+2y^3+2z^3\geq x^2y+xy^2+xz^2+x^2z+yz^2+y^2z\]which can be proven with Muirhead's Inequality (as $[3,0,0]\succ[2,1,0]$ ) or we can add up the symmetric AM-GMs:
\[\dfrac{1}{3}x^2+\dfrac{1}{3}x^2+\dfrac{1}{3}y^2\geq x^2y\]
aops

hbghlyj

artofproblemsolving.com/community/c6h1378979p7635514
arqady wrote:
Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\sqrt{\frac{a}{a+2b+c}}+\sqrt{\frac{b}{b+2c+d}}+\sqrt{\frac{c}{c+2d+a}}+\sqrt{\frac{d}{d+2a+b}}\leq2$$
mudok wrote:
By C-S:
$$(LHS)^2\le \sum_{cyc}a(b+2c+d) \sum_{cyc}\frac{1}{(a+2b+c)(b+2c+d)}$$
$$\sum_{cyc}a(b+2c+d) \sum_{cyc}\frac{1}{(a+2b+c)(b+2c+d)}\le 4 \ \ \iff \ \ (a-c)^2(b-d)^2\ge 0$$
artofproblemsolving.com/community/c6h1377496p7620984

色k

zhihu.com/question/454727066

hbghlyj

令$x=\sqrt{a\over a + k b + c}$, 使用这帖的方法,\[
\det\begin{pmatrix}
1-x^{-2} &k & 1 \\
1 & 1-y^{-2} & k \\
k & 1 & 1-z^{-2}
\end{pmatrix}=0\]即$$-1 +\sum x^2+(k-1)\sum x^2 y^2+ (2-3k+k^3) x^2 y^2 z^2=0$$

1. Maximize[{x+y+z,-1+x^2+y^2-x^2 y^2+k x^2 y^2+z^2-x^2 z^2+k x^2 z^2-y^2 z^2+k y^2 z^2+2 x^2 y^2 z^2-3 k x^2 y^2 z^2+k^3 x^2 y^2 z^2==0},{x,y,z}]

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可知, 当$k≤α≈3.29531$时有不等式\[\sum\sqrt{a\over a + k b + c}\le3\sqrt{1\over2+k}\]
$α$的精确值为Root[100 - 4130*#1 - 2291*#1^2 + 1825*#1^3 - 281*#1^4 + 16*#1^5 & , 3, 0]
当$k>α$时, 最大值满足$x=y≠z$, 比如

1. k=3.3;
2. Maximize[{x+y+z,-1+x^2+y^2-x^2 y^2+k x^2 y^2+z^2-x^2 z^2+k x^2 z^2-y^2 z^2+k y^2 z^2+2 x^2 y^2 z^2-3 k x^2 y^2 z^2+k^3 x^2 y^2 z^2==0},{x,y,z}]

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结果是当x=y=0.403734,z=0.495654时取等.