<s>[INST] 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?
 [/INST] 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.
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