If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$

Question

If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$

The solution goes as follows:

$a=\frac{x}{y+z}$, $b=\frac{y}{z+x}$, $c=\frac{z}{x+y}$. We can see that:

$ab+bc+ac+2abc=1$

It's enough if: $(a+b+c)^2\ge 4-14abc$

$(a+b+c)^2\ge 4(ab+bc+ab+2abc)-14abc$

From Schur it is enough if: $6abc\ge\frac{9abc}{a+b+c}$ which is true from Nesbit.

Could you please provide a more intuitive and easier approach?

Answer 1

if $x,y,z > 0 $, say $a = \frac{x}{y+z} , b=\frac{y}{x+z} , c=\frac{z}{x+y} $ then we'll proof that $$(a+b+c)^2 \ge 4 - 14\cdot a \cdot b \cdot c $$ $(a+b+c)^2 = a^2+2.a.b+b^2+2.a.c+2.b.c+c^2 = a^2+b^2+c^2+2(a.b+b.c+a.c) $

as you said $a.b+a.c+b.c+2.a.b.c = 1$

$a^2+b^2+c^2 +2(1-2.a.b.c) \ge 4 - 14.a.b.c $

$ a^2+b^2+c^2 +2 - 4.a.b.c \ge 4 - 14.a.b.c $

$ a^2+b^2+c^2 +10.a.b.c \ge 2 $

since $x,y,z > 0 $ means that $x$ or $y$ or $z$ are positive numbers, their lowest value is $ 0 + 10^{-n} $ and since our inequality speaks also minimum value, we can get the result by assuming the minimum value of $x,y,z$ there

for the simplest case scenario, this happens when $ x= y= z$ at the lowest level

$ x^2/(z+y)^2+y^2/(z+x)^2+z^2/(y+x)^2 + (10.x.y.z)/((y+x)(z+x)(z+y)) \ge 2 $

so if i say $x=y=z$, the result is $2$ this means that even if $x , y , z$ was $ < 0 $ our inequality would still be greater than 2

Answer 2

Version of 01.04.2021.

Since the task is homogenious, one may denote $$s=x+y+z=3,\quad r=xy+yz+zx,\quad p=xyz,\tag1$$ wherein $$r\le\dfrac13s^2=3,\quad p\le\dfrac1{27}s^3=1. \tag2$$ At the same time, $$p=(3-u)v,\quad r=(3-u)u+v,\tag3$$ where $$u=x+y\in[0,3],\quad v=xy\in\left[0,\dfrac{u^2}4\right].\tag4$$

The given inequality takes the forms of $$4\big(x(s-y)(s-z)+y(s-z)(s-x)+z(s-x)(s-y)\big)^2$$ $$ \ge \big(16(s-x)(s-y)(s-z)-56xyz\big)(s-x)(s-y)(s-z)$$ $$ = \big(4(s-x)(s-y)(s-z)-7xyz\big)^2 -49(xyz)^2,$$ or $$4(s^3-2sr+3p)^2+49p^2\ge (4sr-11p)^2,\tag5$$ wherein, taking in accouint $(1)-(4),$

$$4(s^3-2sr+3p)^2+49p^2-(4sr-11p)^2$$ $$=4(27-6r+3p)^2+49p^2-(12r-11p)^2$$ $$=12(-108r+10pr-3p^2+54p+243)=12f(u,v),$$ $$f(u,v)=243-324u+108u^2+54v+36uv-60u^2v+10u^3v+3v^2+8uv^2-3u^2v^2,$$ $$f'_v(u,v)=54+36u-60u^2+10u^3+6v+16uv-6u^2v.$$ Then at the bounds by $\;v\;$

• $f(u,0)=27(3-2u)^2\ge0,$
• $f\left(u,\dfrac{u^2}4\right)=\dfrac3{16}(1296-1728u+648u^2+48u^3-79u^4+16u^5-u^6)$ $=\dfrac3{16}(9-u^2)(u^2-8u+12)^2\ge0.$

At the stationary points $$2f(u,v)=2f(u,v)-v\,f'_v(u,v)=g(u,v),$$ where

• $g(u,v)=486-648u+216u^2+(54+36u-60u^2+10u^3)v,$
• $g(u,0)=54(3-2u)^2\ge0,$
• $g\left(u,\dfrac{u^2}4\right)=\dfrac12(972-1296u+459u^2+18u^3-30u^4+5u^5)$ $=\dfrac52\left(u+\dfrac{58}{15}\right)\left(u-\dfrac85\right)^2\left(u-\dfrac{10}3\right)^2 +\dfrac{253}{15}\left(u+3\right)\left(u-\dfrac53\right)^2$ $+\dfrac1{1350}\left(95150-119630u+38207u^2\right)\ge0.$

Since $\;g(u,v)\;$ is a linear function by $\;v,\;$ it achives the least value at the edges of the area.

Proved!

Answer 3

Well it's an attempt to delete the square root .

My only idea is to use Bernoulli's inequality and play with the coefficient $\sqrt{2}$.We have :

$$\sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}\leq \sqrt{\alpha}\Big(1+\frac{1}{2}\Big(\frac{4}{\alpha}-1-\frac{14}{\alpha}\frac{xyz}{(x+y)(y+z)(x+z)}\Big)\Big)$$

Remains to show (using the OP's substitution) when:

$$ \sqrt{\alpha}\Big(1+\frac{1}{2}\Big(\frac{4}{\alpha}-1-\frac{14}{\alpha}abc\Big)\Big)\leq a+b+c$$

Wich seems to be easier I think.

The last inequality is equivalent to :

$$\frac{y}{2}+\frac{2}{y}\leq a+b+c+\frac{7}{y}abc$$

With $y^2=\alpha$ so now a bit of algebra to get :

$$y^2+4\leq 2y(a+b+c)+14abc$$

Now putting $y=a+b+c$ we get the inequality of the OP.

The end is the same as the OP.