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.
 A: The required inequality is trivialized by the claim below.  The equality case is when $a=b=c=1$.

Claim: If $a,b,c>0$ are such that $abc=1$, then $\displaystyle\sqrt{\frac{a}{a+8}}\geq \frac{a^{4/9}}{a^{4/9}+b^{4/9}+c^{4/9}}$.  The equality holds if and only if $a=b=c=1$.

Proof: Note that the required inequality is equivalent to
$$\left(a^{4/9}+b^{4/9}+c^{4/9}\right)^2 \geq a^{-1/9}(a+8)\,,$$
which is also equivalent to
$$\left(b^{4/9}+c^{4/9}\right)\left(a^{4/9}+a^{4/9}+b^{4/9}+c^{4/9}\right) \geq 8a^{-1/9}\,.$$
To prove the previous inequality, we invoke the AM-GM Inequality twice:
$$b^{4/9}+c^{4/9}\geq 2b^{2/9}c^{2/9}$$
and
$$a^{4/9}+a^{4/9}+b^{4/9}+c^{4/9}\geq 4a^{1/9}a^{1/9}b^{1/9}c^{1/9}=4a^{2/9}b^{1/9}c^{1/9}\,.$$
Thus, 
$$
\begin{align}
\left(b^{4/9}+c^{4/9}\right)\left(a^{4/9}+a^{4/9}+b^{4/9}+c^{4/9}\right)  &\geq \left(2b^{2/9}c^{2/9}\right)\left(4a^{2/9}b^{1/9}c^{1/9}\right)
\\
&=8a^{2/9}b^{1/3}c^{1/3}=8a^{-1/9}\left(abc\right)^{1/3}=8a^{-1/9}\,,
\end{align}$$
which is what we want.  By the equality condition of the AM-GM Inequality, the equality happens iff $a=b=c=1$.
P.S.:  I just realized why this inequality looks so familiar.  It is equivalent to IMO'2001#2 (http://imo.wolfram.com/problemset/IMO2001_solution2.html).  Substitute $a$, $b$, and $c$ by $x^3$, $y^3$, and $z^3$, then homogenize the required inequality via the condition $xyz=1$, and you will see what I'm talking about.
A: Let $a=\frac{x^2}{yz}$, $b=\frac{y^2}{xz}$ and $c=\frac{z^2}{xy}$, where $x$, $y$ and $z$ are positives.
Hence, by Holder and AM-GM we obtain:
$$\sum_{cyc}\sqrt{\frac{a}{a+8}}=\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{x}{\sqrt{x^2+8yz}}\right)^2\sum\limits_{cyc}x(x^2+8yz)}{\sum\limits_{cyc}x(x^2+8yz)}}\geq$$
$$\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}(x^3+8xyz)}}\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}(x^3+3x^2y+3x^2z+2xyz)}}=1.$$
Done!
A: 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!
A: Replace $(a,b,c) \to \left(\frac{a^2}{bc},\frac{b^2}{ca},\frac{c^2}{ab}\right)$ the inequality become
$$ \sum \frac{a}{\sqrt{a^2+8bc}} \geqslant 1.$$
We have
$$\left[5(a^2+b^2+c^2)+4(ab+bc+ca)\right]^2-(a^2+8bc)(5a+2b+2c)^2$$
$$=[5a(b+c)+17bc](2a-b-c)^2+[82a^2+25(b^2+c^2)+35a(b+c)+41bc](b-c)^2 \geqslant 0.$$
Therefore
$${\frac {a}{\sqrt {a^{2} + 8bc}}}\geq {\frac {a(5a + 2b + 2c)}{5\left(a^2 + b^2 + c^2\right) + 4(bc + ca + ab)}}.$$
Note
$$\sum{\frac {a(5a + 2b + 2c)}{5\left(a^2 + b^2 + c^2\right) + 4(ab+bc+ca)}} = 1,$$
The proof is completed.
A: First let
$$ x = \sqrt{\frac{a}{a+8}}, \,\,  y = \sqrt{\frac{b}{b+8}}, \,\,  z = \sqrt{\frac{c}{c+8}} \,\, $$
Then $1 > x,y,z > 0$ and 
$$ a = \frac{8x^2}{1 - x^2}, \,\,   b = \frac{8y^2}{1 - y^2}, \,\,  c = \frac{8z^2}{1 - z^2},\,\, $$
So the question transforms to this:
Given that $1 > x,y,z > 0, \, \, \frac{512x^2y^2z^2}{(1 - x^2)(1 - y^2)(1 - z^2)} = 1$, prove that $x + y + z \geqslant 1$.
Prove this by contradiction. Suppose on the contrary that $x + y + z < 1$, then 
$$ \begin{align}
(1 - x^2)(1 - y^2)(1 - z^2) &= (1 - x)(1 + x)(1 - y)(1 + y)(1 - z)(1 + z) \\
&>(x + x + y + z)(y + z)(x + y + y + z)(x + z)(z + x + y + z)(x + y) \\
&\geqslant 4x^{\frac12}y^{\frac14}z^{\frac14}\cdot 2y^{\frac12}z^{\frac12} \cdot 4y^{\frac12}x^{\frac14}z^{\frac14}\cdot 2x^{\frac12}z^{\frac12} \cdot 4z^{\frac12}y^{\frac14}x^{\frac14}\cdot 2y^{\frac12}x^{\frac12}\\
&=512 x^{\frac12 + \frac14 + \frac12 + \frac14 + \frac12}y^{\frac14 + \frac12 + \frac12 + \frac14 + \frac12}z^{\frac14 + \frac12 +\frac14 + \frac12 + \frac12} \\
&= 512x^2y^2z^2
\end{align}$$
And this is contradictory to the condition.
A: Let $a=\frac{x^2}{yz},\, b=\frac{y^2}{zx},\,c=\frac{z^2}{xy}.$ Thus$,$ we need to prove$:$
$$\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8zx}} +\frac{z}{\sqrt{z^2+8xy}} \geqq 1$$
By AM-GM$:$ \begin{align*} \text{LHS} &=\sum\limits_{cyc} \frac{x}{\sqrt{x^2+8yz}} =\sum\limits_{cyc} \frac{x(x+y+z)}{\sqrt{(x^2+8yz)(x+y+z)^2}}\\&\geqq 2\sum\limits_{cyc} \frac{x(x+y+z)}{(x^2+8yz)+(x+y+z)^2} \geqq 1 \end{align*}
Where the last inequality equivalent to $$\frac{1}{2} \sum\limits_{cyc} \left( 8\,{x}^{3}y+31\,{x}^{2}{y}^{2}+8\,x{y}^{3}+202\,x{y}^{2}z+262
\,xy{z}^{2}+202\,x{z}^{3}+79\,{z}^{4} \right)  \left( x-y \right) ^{2}
 \geqq 0$$
Done.
