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?
 A: 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
A: 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.
