Prove that $\sum\limits_{\mathrm{cyc}}{\frac{1}{(x+2y)^2}} \geq\frac{1}{xy+yz+zx}$ for $x, y, z > 0$

Question

Let $x,y,z>0$. Prove that $$\frac{1}{(x+2y)^2}+\frac{1}{(y+2z)^2}+\frac{1}{(z+2x)^2} \geq\frac{1}{xy+yz+zx}.$$

I tried to apply Cauchy - Schwarz's inequality but I couldn't prove this inequality!

Answer 1

$a=x+2y,b=y+2z,c=z+2x \implies x=\dfrac{a-2b+4c}{9},y=\dfrac{b-2c+4a}{9},z=\dfrac{c-2a+4b}{9}$

the inequality become:

$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \ge \dfrac{27}{5(ab+bc+ac)-2(a^2+b^2+c^2)}$

now use UVW method:

$3u=a+b+c,3v^2=ab+bc+ac,w^3=abc \implies u\ge v\ge w$

$\iff \dfrac{(3v^2)^2-6uw^3}{w^6}\ge \dfrac{3}{3v^2-2u^2} \iff w^6+2u(3v^2-2u^2)w^3-3v^4(3v^2-u^2) \le 0$

let $w^3=x,f(x)=x^2+2u(3v^2-2u^2)x-3v^4(3v^2-2u^2)$

$2u(3v^2-2u^2) \ge 0 $

$f_{max}(x)=f(w^3|w=v)=f(v^3)$, when $w=v \implies u=v=w \implies f(v^3)=0 \implies f(x) \le 0 $

when $u=v=w \implies a=b=c \implies x=y=z$

QED.

Answer 2

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2$.

Hence, $t\geq1$ and we need to prove that $$\sum_{cyc}(4a^5b+4a^4c-12a^4b^2+12a^4c^2+5a^3b^3+8a^4bc-19a^3b^2c+5a^3c^2b-7a^2b^2c^2)\geq0$$ or $$\sum_{cyc}(4a^5b+4a^4c+5a^3b^3+8a^4bc-7a^3b^2c-7a^3c^2b-7a^2b^2c^2)\geq$$ $$\geq12\sum_{cyc}(a^4b^2-a^4c^2+a^3b^2c-a^3c^2b)$$ or $$\sum_{cyc}(4a^5b+4a^4c+5a^3b^3+8a^4bc-7a^3b^2c-7a^3c^2b-7a^2b^2c^2)\geq$$ $$\geq12(a-b)(a-c)(b-c)(a+b+c)(ab+ac+bc).$$ Since by Schur $3\sum\limits_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0$, it's enough to prove that $$\sum_{cyc}(4a^5b+4a^4c+2a^3b^3+8a^4bc-4a^3b^2c-4a^3c^2b-10a^2b^2c^2)\geq$$ $$\geq12(a-b)(a-c)(b-c)(a+b+c)(ab+ac+bc)$$ or $$\sum_{cyc}(2a^5b+2a^4c+a^3b^3+4a^4bc-2a^3b^2c-2a^3c^2b-5a^2b^2c^2)\geq$$ $$\geq6(a-b)(a-c)(b-c)(a+b+c)(ab+ac+bc)$$ or $$18u^4v^2-24u^2v^4+5v^6+2u^3w^3-uv^2w^3\geq2(a-b)(a-c)(b-c)uv^2$$ and since $18u^4v^2-24u^2v^4+5v^6+2u^3w^3-uv^2w^3\geq0$, it's enough to prove that $$\left(18u^4v^2-24u^2v^4+5v^6+2u^3w^3-uv^2w^3\right)^2\geq4(a-b)^2(a-c)^2(b-c)^2u^2v^4$$ or $$\left(18u^4v^2-24u^2v^4+5v^6+2u^3w^3-uv^2w^3\right)^2\geq108(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)u^2v^4$$ or $$(4u^4-4u^2v^2+109v^4)u^2w^6+2uv^2(36u^6+150u^4v^2-290u^2v^4-5v^6)w^3+$$ $$+(324u^8-864u^6v^2+432u^4v^4+192u^2v^6+25v^8)v^4\geq0.$$ Since $324t^4-864t^3+432t^2+192t+25>0$, it's remains to prove that $$t(36t^3+150t^2-290t-5)^2-t(4t^2-4t+109)(324t^4-864t^3+432t^2+192t+25)\leq0$$ or $$t(t-1)^2(144t^3-72t^2-216t-25)\leq0,$$ where $t\geq1$ and $36t^3+150t^2-290t-5<0$.

Let $p(t)=36t^3+150t^2-290t-5$ and $q(t)=144t^3-72t^2-216t-25$.

By Descartes $p$ has an unique positive root $x_1$ and $q$ has an unique positive root $x_2$.

But $p(1.5)=19>0$ and $q(1.5)=-25<0$, which says that $q(x)<0$ for all $1\leq x\leq x_1$

and we are done!

Answer 3

Also we can make the following.

From my previous post we need to prove that: $$\sum_{cyc}(4a^5b+4a^4c-12a^4b^2+12a^4c^2+5a^3b^3+8a^4bc-19a^3b^2c+5a^3c^2b-7a^2b^2c^2)\geq0$$ or $$6\sum_{cyc}ab(a^2-b^2-2ab+2ac)^2+\sum_{sym}(2a^5b-a^3b^3-4a^4bc+10a^3b^2c-7a^2b^2c^2)\geq0,$$ which is obviously true.

Answer 4

Assume $z=min\{x,y,z\}$, we prove two inequalities $$$$1. $$\sum_{cyc} \frac{1}{(2x+y)^2} \geqslant \frac{1}{9xy}+\frac{8}{9(x+z)(y+z)}$$ which is true by expanding $$$$2.$$\frac{1}{9xy}+\frac{8}{9(x+z)(y+z)} \geqslant \frac{1}{xy+yz+zx}$$ which is equivalent to $(x+y)(y+z)(z+x) \geqslant 8xyz$

Answer 5

Alternative proof without full expanding or uvw:

Let $$u = \frac{1}{x + 2y}, \quad v = \frac{1}{y + 2z}, \quad w = \frac{1}{z + 2x}.$$

We have \begin{align*} &\mathrm{LHS} - \mathrm{RHS} \\ =\,& u^2 + v^2 + w^2 - \frac{1}{xy + yz + zx}\\ =\,& \frac{(u-v)^2 + (v - w)^2 + (w-u)^2}{2} + uv + vw + wu - \frac{1}{xy + yz + zx}\\ =\,& \frac{(u-v)^2 + (v - w)^2 + (w-u)^2}{2} + \frac{3(x + y + z)}{(x + 2y)(y + 2z)(z + 2x)} - \frac{1}{xy + yz + zx}\\ =\,& \frac{(u-v)^2 + (v - w)^2 + (w-u)^2}{2} - \frac{(y-x)(y-z)(x-z)}{(x+2y)(y+2z)(z+2x)(xy+yz+zx)}. \tag{1} \end{align*}

WLOG, assume that $z = \min(x, y, z)$.

From (1), we only need to prove the case $y \ge x \ge z$.

If $2x \ge y + z$, from (1), we have \begin{align*} \mathrm{LHS} - \mathrm{RHS} &\ge \frac{(v - u + v - w + w - u)^2}{6} - \frac{(y-x)\cdot (y - z + x - z)^2/4}{(x+2y)(y+2z)(y+2z)xy} \tag{2}\\ &= \frac{2(x + y - 2z)^2}{3(x + 2y)^2(y + 2z)^2} - \frac{(y-x) (x + y - 2z)^2}{4(x+2y)(y+2z)^2xy}\\ &= \frac{(x + y - 2z)^2}{(x+2y)(y+2z)^2} \left(\frac{2}{3(x + 2y)} - \frac{y-x}{4xy}\right)\\ &\ge 0 \end{align*} where in (2) we use $(u-v)^2 + (v - w)^2 + (w-u)^2 \ge (v-u + v - w + w - u)^2 /3$, and $(y - z)(x - z) \le (y - z + x - z)^2/4$, and $z + 2x \ge y + 2z$, and $xy + yz + zx \ge xy$; and the last inequality follows from $\frac{2}{3(x + 2y)} - \frac{y-x}{4xy} \ge \frac{2}{3(x + 4x)} - \frac{y-x}{4xy} = \frac{15x -7y}{60xy} \ge 0$.

If $2x < y + z$, from (1), we have \begin{align*} \mathrm{LHS} - \mathrm{RHS} &\ge \frac{(v - u + w - v + w - u)^2}{6} - \frac{(x-z)(2y - x - z)^2/4}{(x+2y)(z+2x)(z+2x)xy} \tag{3}\\ &= \frac{2(2y - x - z)^2}{(x+2y)^2(z+2x)^2} - \frac{(x-z)(2y - x - z)^2}{4(x+2y)(z+2x)^2xy}\\ &= \frac{(2y - x - z)^2}{(x + 2y)(z + 2x)^2} \left(\frac{2}{3(x + 2y)} - \frac{x - z}{4xy}\right)\\ &\ge 0 \end{align*} where in (3) we use $(u-v)^2 + (v-w)^2 + (w-u)^2 \ge (v - u + w - v + w - u)^2/3$, and $(y - x)(y - z) \le (y - x + y - z)^2/4$, and $y+2z \ge z + 2x$, and $xy + yz + zx \ge xy$; and the last inequality follows from $\frac{2}{3(x + 2y)} - \frac{x - z}{4xy} = \frac{2xy + 3zx + 6yz - 3x^2}{12(x+2y)xy} \ge 0 $ (using $2xy + 2zx = 2x(y + z) \ge 2x \cdot 2x = 4x^2$).

We are done.