Prove: 0 < a < 1 < b < 3 < c < 4 | POTW #436 (Sept. 28, 2020)

[anemone]

Here is this week's POTW:

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Let $a<b<c$ be real numbers such that $a+b+c=6$ and $ab+bc+ca=9$. Prove that $0<a<1<b<3<c<4$.

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[anemone]

Hi all!

Just to let you all know that I will extend the deadline to solve last week's POTW (which actually due today) to next week, with the great hope to receive any submission of solution from the members! (Happy)

 

[anemone]

Spoiler: Suggested solution of other

Let $abc=k$, this makes $a,\,b$ and $c$ become the roots of the cubic equation $x^3-6x^2+9x-k=0$.

The cubic equation can be rewritten as $x(x-3)^2=k$.

Plot the graph of $f(x)=x(x-3)^2$ and the horizontal line of $y=k$ with condition $0<k<4$ on the same diagram. Since the roots are real, the line $y=k$ can only be moved in such a way that it always cuts the curve at 3 distinct points in such a way that $0<a<1<b<3<c<4$.

[TIKZ][scale=3]
\draw[help lines] (0,0) grid (5,4);
\draw[thick,->] (0,0) -- (4.5,0) node[anchor=north west] {x axis};
\draw[thick,->] (0,0) -- (0,4.5) node[anchor=south east] {y axis};
\draw[ domain=-0.2:4, samples=100] plot (\x,\x^3 - 6*\x^2 +9*\x);
\draw (0,3.2) -- (4,3.2);
\foreach \x in {0,1,2,3,4}
\draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {$\x$};
\foreach \y in {0,1,2,3,4}
\draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {$\y$};
\node at (3.6,2) {\large $y=f(x)$};
\node at (4.2,3.2) {\large $y=k$};
\draw[ densely dashed,color=blue] (0,4) -- (4,4);
\draw[ densely dashed,color=blue] (4,0) -- (4,4);
\draw[ densely dashed,color=blue] (0.52,0) -- (0.52,3.2);
\draw[ densely dashed,color=blue] (1,0) -- (1,3.2);
\draw[ densely dashed,color=blue] (1.58,0) -- (1.58,3.2);
\draw[ densely dashed,color=blue] (3.9,0) -- (3.9,3.2);
\node at (0.52,-0.08) {$a$};
\node at (1.58,-0.08) {$b$};
\node at (3.9,-0.08) {$c$};
[/TIKZ]

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