Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$ If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the reverse inequality). Need some stronger inequality. Thanks.

Let $a=\frac{x^2}{yz}$ and $b=\frac{y^2}{xz}$, where $x$, $y$ and $z$ are positives. 
Hence, $c=\frac{z^2}{xy}$ and we need to prove that:
$$\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}\geq1.$$
Now, by Holder
$$\left(\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}\right)^2\sum_{cyc}x(x^2+8yz)\geq(x+y+z)^3.$$
Thus, it remains to prove that
$$(x+y+z)^3\geq\sum_{cyc}x(x^2+8yz)$$ or
$$\sum_{cyc}z(x-y)^2\geq0.$$
Done!
Another way:
$$\sum_{cyc}\sqrt{\frac{a}{a+8}}=\sum_{cyc}\frac{x}{\sqrt{x^2+8xy}}\geq\sum_{cyc}\frac{x^{\frac{4}{3}}}{x^{\frac{4}{3}}+y^{\frac{4}{3}}+z^{\frac{4}{3}}}=1$$
Also we can use the Contradiction method.
Let $\frac{a}{a+8}=\frac{p^2}{9}$, $\frac{b}{b+8}=\frac{q^2}{9}$ and $\frac{c}{c+8}=\frac{q^2}{9}$, where $p$, $q$ and $r$ are positives.
Hence, we need to prove that $p+q+r\geq3.$
But the condition $abc=1$, gives
$$8^3=\prod_{cyc}\left(\frac{9}{p^2}-1\right)$$ or
$$81=57p^2q^2r^2-p^2q^2-p^2r^2-q^2r^2+9(p^2+q^2+r^2).$$
Now, let $p+q+r<3$, $p=kx$, $q=y$ and $r=z$, where $k>0$ and $x+y+z=3$.
Hence, $kx+y+z<3=x+y+z$, which gives $0<k<1$.
Thus, since $9-p^2-q^2>(p+q)^2-p^2-q^2>0$, we obtain:
$$81=57p^2q^2r^2-p^2q^2-p^2r^2-q^2r^2+9(p^2+q^2+r^2)=$$
$$=k^2x^2(57y^2z^2-y^2-z^2+9)-y^2z^2+9(y^2+z^2)<$$
$$<x^2(57y^2z^2-y^2-z^2+9)-y^2z^2+9(y^2+z^2),$$
which is contradiction because we'll prove now that
$$57x^2y^2z^2-x^2y^2-x^2z^2-y^2z^2+9(x^2+y^2+z^2)\leq81.$$
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, it's obvious that our inequality is equivalent to $f(w^3)\leq0$, where $f$ is a convex function.
Id est, it's enough to prove the last inequality for an extremal value of $w^3$,
which happens in the following cases.


*

*$y=x$, $z=3-2x$, where $0<x<1.5$, which gives
$$x(x-1)^2(9+15x+19x^2-19x^3)\geq0,$$
which is true for $0<x<\frac{3}{2}$;

*$w^3=0$.
Let $z=0$ and $y=3-x$, where $0<x<3$.
We obtain, $x(4x^3-12x^2-9x+54)\geq0$, which is obvious.
Done again!
