Ellie Benjamin

2022-07-05

Prove $\sum \frac{1}{\left(x+2y{\right)}^{2}}\ge \frac{1}{xy+yz+zx}$
Let $x,y,z>0$ Prove that:
$\frac{1}{\left(x+2y{\right)}^{2}}+\frac{1}{\left(y+2z{\right)}^{2}}+\frac{1}{\left(z+2x{\right)}^{2}}\ge \frac{1}{xy+yz+zx}$
I tried to apply Cauchy - Schwarz's inequality but I couldn't prove this inequality!

$a=x+2y,b=y+2z,c=z+2x\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=\frac{a-2b+4c}{9},y=\frac{b-2c+4a}{9},z=\frac{c-2a+4b}{9}$
the inequality become:
$\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}+\frac{1}{{c}^{2}}\ge \frac{27}{5\left(ab+bc+ac\right)-2\left({a}^{2}+{b}^{2}+{c}^{2}\right)}$
now use UVW method:
$3u=a+b+c,3{v}^{2}=ab+bc+ac,{w}^{3}=abc\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}u\ge v\ge w$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\frac{\left(3{v}^{2}{\right)}^{2}-6u{w}^{3}}{{w}^{6}}\ge \frac{3}{3{v}^{2}-2{u}^{2}}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{w}^{6}+2u\left(3{v}^{2}-2{u}^{2}\right){w}^{3}-3{v}^{4}\left(3{v}^{2}-{u}^{2}\right)\le 0$
let ${w}^{3}=x,f\left(x\right)={x}^{2}+2u\left(3{v}^{2}-2{u}^{2}\right)x-3{v}^{4}\left(3{v}^{2}-2{u}^{2}\right)$
$2u\left(3{v}^{2}-2{u}^{2}\right)\ge 0$
${f}_{max}\left(x\right)=f\left({w}^{3}|w=v\right)=f\left({v}^{3}\right)$, when
$w=v\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}u=v=w\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}f\left({v}^{3}\right)=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}f\left(x\right)\le 0$
when $u=v=w\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}a=b=c\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=y=z$
QED.

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