MHB Can You Prove This Inequality with Positive Real Numbers?

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The problem presented requires proving the inequality involving positive real numbers \(a\), \(b\), and \(c\) under the condition \(abc=1\). The inequality states that \(\sqrt{\dfrac{a}{a+8}}+\sqrt{\dfrac{b}{b+8}}+\sqrt{\dfrac{c}{c+8}}\ge 1\). Although no participants provided a solution in the discussion, references to external solutions are suggested for further exploration. The challenge lies in applying appropriate mathematical techniques to validate the inequality. Engaging with this problem can enhance understanding of inequalities in the context of positive real numbers.
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Here is this week's POTW:

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If $a,\,b$ and $c$ are positive reals such that $abc=1$, prove that $\sqrt{\dfrac{a}{a+8}}+\sqrt{\dfrac{b}{b+8}}+\sqrt{\dfrac{c}{c+8}}\ge 1$.

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No one answered this POTW. However, you can refer to the solution from other as shown below.

Let $x=\sqrt{\dfrac{a}{a+8}},\,y=\sqrt{\dfrac{b}{b+8}},\,\,z=\sqrt{\dfrac{c}{c+8}}$ then we have $1>x,\,y,\,z>0$ and $a=\dfrac{8x^2}{1-x^2},\,b=\dfrac{8y^2}{1-y^2},\,c=\dfrac{8z^2}{1-z^2}$.

We are essentially asking to prove $x+y+z\ge 1$, given $1>x,\,y,\,z>0$ and $\dfrac{512x^2y^2z^2}{(1-x^2)(1-y^2)(1-z^2)}=1$.

The plan is to prove it by contradiction.

Suppose on the contrary that $x+y+z<1$, then

$\begin{align*}(1-x^2)(1-y^2)(1-z^2)&=(1-x)(1+x)(1-y)(1+y)(1-z)(1+z)\\&>((x+x+y+z)(y+z)((x+y+y+z)(x+z)((z+x+y+z)(x+y)\\& \ge 4x^{\frac{1}{2}}y^{\frac{1}{4}}z^{\frac{1}{4}}\cdot 2y^{\frac{1}{2}}z^{\frac{1}{2}}\cdot 4y^{\frac{1}{2}}x^{\frac{1}{4}}z^{\frac{1}{4}}\cdot 2x^{\frac{1}{2}}z^{\frac{1}{2}}\cdot 4z^{\frac{1}{2}}y^{\frac{1}{4}}x^{\frac{1}{4}} \cdot 2y^{\frac{1}{2}}x^{\frac{1}{2}}\\&=512x^{\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}}y^{\frac{1}{4}+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}}z^{\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{2}}\\&=512x^2y^2z^2\end{align*}$

And this leads to $\dfrac{512x^2y^2z^2}{(1-x^2)(1-y^2)(1-z^2)}\le1$, a contradiction and therefore the proof follows.
 
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