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- Feb 14, 2012

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- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,937

- Thread starter
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- #2

- Feb 14, 2012

- 3,937

My solution:

$f(t)=(t-a)(t-b)(t-c)=t^3-(a+b+c)t^2+(ab+bc+ca)t-abc$ and

$g(t)=(t-x)(t-y)(t-z)=t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz$

Then we see that:

$f(t)-g(t)=(ab+bc+ca-xy-yz-zx)t$ is a straight line.

At $t=z$, $f(t)-g(t)=f(z)-g(z)=f(z)<0$

Also, when $t=0$, we have

$\begin{align*}f(t)-g(t)&=f(0)-g(0)\\&=-abc-(-xyz)\\&=-abc+xyz\\&=0 \end{align*}$

Hence we can draw conclusion that the straight line $f(t)-g(t)=(ab+bc+ca-xy-yz-zx)t$ is always negative for $t>0$.

That is, $f(t)-g(t)<0$ or $f(t)<g(t)$ for $0<t<a$. This holds only when $x \ge a$.

Hence we're done.

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- Feb 14, 2012

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I realized only by now that I've posted the terribly wrong answer to this particular challenge problem of mine and I am extremely sorry about that.

I will now post another suggested solution by other and please discard my previous solution post, thanks.

$f(t)=(t-a)(t-b)(t-c)=t^3-(a+b+c)t^2+(ab+bc+ca)t-abc$ and

$g(t)=(t-x)(t-y)(t-z)=t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz$

Then we see that:

$f(t)-g(t)=(ab+bc+ca-xy-yz-zx)t$ never changes sign for positive values of $t$.

Since $f(t)>0$ for $t>c$, we have that $f(z)-g(z)=f(z) \ge 0$, so that $f(t) \ge g(t)$ for all $t>0$.

Hence, for $0<t<a$, $g(t) \le f(t) <0$, from which it follows that $g(t)$ has no root less than $a$. Hence $x \ge a$ as desired.

- Jan 25, 2013

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- Feb 14, 2012

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Hican you find another sets of values ?

(a<b<c , x<y<z and c<z )

(a+b+c=x+y+z)

(abc=xyz)

with a,b,c,x,y,z>0

Here is another set of the values which is not very hard to find:

$(x,\,y,\,z)=(5,\,5.2,\,8.1)$

$(a,\,b,\,c)=(4.5,\,6,\,7.8)$