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

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?

• So are you facing difficulty on how they got $ab+bc+ca+2abc=1$?there is anice proof for that Mar 6 at 17:40
• @AlbusDumbledore actually that's exactly what is giving me difficulties, I can't prove it Mar 6 at 17:43
• @AlbusDumbledore sorry for having deleted the chat, it's the first time I created a chat and I thought I had done something wrong. Sorry about that. Thank you so much for your response, your reasoning is brilliant, I have understood it implicitly. Have a great day Mar 6 at 18:53
• @MichaelBlane We can get $a + b + c \ge \frac{3}{2}$ from $ab+bc+ac+2abc=1$ without using Nesbitt. Indeed, from $(a + b + c)^2 \ge 3(ab + bc + ca)$ and $a + b + c \ge 3\sqrt[3]{abc}$, we can easy get $a + b + c \ge \frac{3}{2}$. Mar 9 at 2:04
• @V.S.e.H. I think it is fine as comment. From $(a + b + c)^2 \ge 3(ab + bc + ca)$ and $a + b + c \ge 3\sqrt[3]{abc}$, we have $\frac{(a + b + c)^2}{3} + 2 \cdot (\frac{a + b + c}{3})^3 \ge ab + bc + ca + 2abc = 1$ which results in $a + b + c \ge 3/2$. Mar 14 at 15:13

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

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.

• @RiverLi Good point but If you read carefully I say that it works only for some values. Mar 14 at 8:28
• @RiverLi Well there is a typo ! Mar 14 at 8:38
• @RiverLi Well you're right I forgot that and maybe it's a bad idea .But I don't see better than the OP's proof ! Mar 14 at 9:44
• @RiverLi I think there is something wrong can you tell me where ? Mar 14 at 10:17
• @RiverLi So do you think it's a nice refinement ? Mar 16 at 10:20

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!

• Why offtopic? I think your solution is wrong. You said "$2p^3+3p^2-1=(p+1)^2(2p-1)\ge 0,\quad p\ge\dfrac12$", it should be $2p^3+3p^2-1 \le 0$. You make a mistake $2p^3+3p^2-1 \ge 0$. Mar 8 at 3:34
• How is this "neater" and "nicer" than the solution provided by OP? Mar 14 at 12:45
• @YuriNegometyanov Check: $4p-5r=5xy(3-x-y)-4\big(xy+(x+y)(3-x-y)\big)$. Mar 14 at 14:10
• @YuriNegometyanov $4p\ge 5r-11$ is incorrect, e.g., for $x = 0, y = z = 3/2$. Or $x = 1/100, y = z = 299/200$. Mar 14 at 14:23
• @YuriNegometyanov Sorry, it is still incorrect. $4(27-6r+3p)^2+49p^2-(12r-11p)^2 = 12(-3p^2+10pr+54p-108r+243)$ is wrong. E.g., for $x = y = z = 1$ (so $s = 3, r = 3, p = 1$), we have $4(27-6r+3p)^2+49p^2-(12r-11p)^2 = 0$, however, $12(-3p^2+10pr+54p-108r+243) = 576$. Mar 15 at 0:57