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Problem Of The Week #482 September 6th 2021

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,937
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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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,937
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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