Prove that for positive reals $a,b,c$, \[\frac{8a^2+2ab}{(b+\sqrt{6ac}+3c)^2}+\frac{2b^2+3bc}{(3c+\sqrt{2ab}+2a)^2}+\frac{18c^2+6ac}{(2a+\sqrt{3bc}+b})^2\geq 1\]
Problem
Source: 2013 Taiwan TST
Tags: Inequality, algebra, Taiwan, Taiwan TST 2013
v_Enhance
12.07.2015 16:14
Letting $x^2=2a$, $y^2=b$, $z^2=3c$, the inequality becomes \[ \sum_{\text{cyc}} \frac{2x^4+y^2z^2}{\left( y^2+z^2+xz \right)^2} \ge 1. \] I did a full expansion of this a couple years ago, but I can't seem to remember if there's a nice way to finish off.
USJL
12.07.2015 16:18
I remember that this is from a problem in Korea TST, but it seems that I fail to find it immediately....
rkm0959
12.07.2015 17:14
Proceed to FKMO 2012 P1
trumpeter
05.01.2016 03:35
v_Enhance wrote: Letting $x^2=2a$, $y^2=b$, $z^2=3c$, the inequality becomes \[ \sum_{\text{cyc}} \frac{2x^4+y^2z^2}{\left( y^2+z^2+xz \right)^2} \ge 1. \]I did a full expansion of this a couple years ago, but I can't seem to remember if there's a nice way to finish off. I believe that it should be $$\sum_{\text{cyc}}\frac{2x^4+x^2y^2}{\left(y^2+z^2+xz\right)^2}\geq1$$instead...
First make the substitution $x=2a$, $y=b$, and $z=3c$. The inequality reduces to $$\sum_{\text{cyc}}\frac{2x^2+xy}{\left(y+\sqrt{xz}+z\right)^2}\geq1.$$By Cauchy-Schwarz, $$\left(y+\sqrt{xz}+z\right)^2\leq\left(y+x+z\right)\left(y+z+z\right),$$so it suffices to prove that $$\frac{1}{x+y+z}\sum_{\text{cyc}}\frac{2x^2+xy}{y+2z}\geq1.$$But $\frac{2x^2+xy}{y+2z}=x\left(\frac{y+2x}{y+2z}\right)=x\left(1+\frac{2x-2z}{y+2z}\right)$, so the LHS is just $$\frac{1}{x+y+z}\left(x+y+z+2\sum_{\text{cyc}}\frac{x^2-xz}{y+2z}\right),$$so it suffices to prove that $$\sum_{\text{cyc}}\frac{x^2-xz}{y+2z}\geq0.$$Multiplying everything by $\left(y+2z\right)\left(z+2x\right)\left(x+2y\right)$ gives that we wish to prove that $$\sum_{\text{cyc}}\left(x^2-xz\right)\left(z+2x\right)\left(x+2y\right)=\sum_{\text{cyc}}2x^4+4x^3y-x^3z-x^2z^2-2x^2yz-2xyz^2\geq0.$$Notice that this inequality is equivalent to $$2x^4+2y^4+2z^4+x^2yz+xy^2z+xyz^2+5x^3y+5y^3z+5xz^3\geq x^3y+x^3z+y^3z+xy^3+xz^3+yz^3+x^2y^2+y^2z^2+x^2z^2+5x^2yz+5xy^2z+5xyz^2.$$But $$x^4+y^4+z^4+x^2yz+xy^2z+xyz^2\geq x^3y+x^3z+y^3z+xy^3+xz^3+yz^3$$by Schur's with $r=2$, $$x^4+y^4+z^4\geq x^2y^2+y^2z^2+x^2z^2$$by AM-GM with $\frac{x^4+y^4}{2}\geq x^2y^2$ and summing cyclically, and $$5x^3y+5y^3z+5xz^3\geq 5x^2yz+5xy^2z+5xyz^2$$by AM-GM with $$\frac{4x^3y+y^3z+2xz^3}{7}\geq x^2yz$$and summing cyclically then multiplying by $5$. Adding these three inequalities together gives the one we want, so the original inequality is true. Equality holds when $x=y=z$, which is when $\left(a,b,c\right)=\left(k,2k,\frac{2}{3}k\right)$ for a positive constant $k$.
Q.E.D.
Mr.A
07.05.2016 13:04
Let $$x=2a,y=b,z=3c$$$$\Rightarrow$$$$\sum_{cyc}\frac{x*(2x+y)}{(y+\sqrt{xz}+z)^2}\geq 1 $$By Cauchy-Schwarz inequality $$(2x+y)(\frac{z^2}{x}+z+y)=(x+x+y)(\frac{z^2}{x}+z+y)\geq (y+\sqrt{xz}+z)^2$$$$\sum_{cyc}\frac{x(2x+y)(\frac{z^2}{x}+z+y)}{(y+\sqrt{xz}+z)^2(\frac{z^2}{x}+z+y)}\geq \sum_{cyc}\frac{x}{\frac{z^2}{x}+z+y}=\sum_{cyc}\frac{x^2}{z^2+xz+xy}$$By Cauchy-Schwarz inequality $$\sum_{cyc}\frac{x^2}{z^2+xz+xy}\geq \frac{(x+y+z)^2}{(x^2+y^2+z^2+2xy+2yz+2zx)}$$Q.E.D
Korean_fish_Kaohsiung
08.07.2024 16:35
Let$$x=2a,y=b,z=3c$$, bacuse the inequality is homogeneous we can let $x+y+z=1$$$\Rightarrow$$$$\sum_{cyc}\frac{x(2x+y)}{(y+\sqrt{xz}+z)^2}\geq 1 $$by AM-GM $$\sum_{cyc}\frac{x(2x+y)}{(y+\sqrt{xz}+z)^2}\geq \sum_{cyc}\frac{x(2x+y)}{(\frac{1}{2}x+y+\frac{3}{2}z)^2}$$while $$\sum_{cyc}\frac{x(2x+y)}{(\frac{1}{2}x+y+\frac{3}{2}z)^2} = \sum_{cyc}\frac{4x(2x+y)}{(3(x+y+z)-(2x+y))^2}=\sum_{cyc}\frac{4x(2x+y)}{(3-(2x+y))^2}$$take the dirivative of the function $f(x)=\frac{x}{(3-x)^2}$ twice to get $f^{''}(x)=\frac{2(x+6)}{(3-x)^4}$ which is always positive when $0<x<3$ so we can apply Jenson's Inequality to get $$4\sum_{cyc}\frac{x(2x+y)}{(3-(2x+y))^2} \geq 4\frac{2x^2+2y^2+2z^2+xy+yz+zx}{(3-(2x^2+2y^2+2z^2+xy+yz+zx))^2}$$and it's easy to know that $f(x) \geq \frac{1}{4}$ when $1 \leq x< 2$ while it's trivial that $$1 = x^2+y^2+z^2+2xy+2yz+2zx \leq 2x^2+2y^2+2z^2+xy+yz+zx < 2x^2+2y+2z^2+4xy+4yz+4zx = 2 $$so $$\sum_{cyc}\frac{x(2x+y)}{(y+\sqrt{xz}+z)^2}\geq f(2x^2+2y^2+2z^2+xy+yz+zx) \geq \frac{1}{4}*4 =1$$and we are done