The answer is D ( equal to 100 )

The answer is D

Sorry ! The solution above is correct when the numbers are natural numbers.I'll solve it again

\(\left\{{}\begin{matrix}a+b+c=20\\c>0\end{matrix}\right.\)\(\Rightarrow a+b< 20\).We have :

\(\dfrac{a+b}{2}\ge\sqrt{ab}\) (AM-GM inequality)

\(\Rightarrow\sqrt{ab}\le\dfrac{a+b}{2}< 10\Rightarrow ab< 100\)

Hence,the answer is D

Let a,b,c be the numbers and assume that a,b are 2 largest numbers

\(\left(a,b,c\in Z^+\right)\)

We have : \(a>c;b>c\Rightarrow a+b>2c\Rightarrow a+b+c>3c\)

\(\Rightarrow3c< 20\Rightarrow c\le6\Rightarrow a+b\ge14\) but \(c\ge1\Rightarrow a+b\le19\)

So,\(14\le a+b\le19\).We have :

\(\dfrac{a+b}{2}\ge\sqrt{ab}\) (AM-GM inequality)

\(\Rightarrow\dfrac{19}{2}\ge\dfrac{a+b}{2}\ge\sqrt{ab}\Rightarrow ab\le\dfrac{361}{4}=90.25< 99< 100\)

\(c\ge1\Rightarrow a>1;b>1\Rightarrow ab>1>0.001\)

If ab = 25,then (a ; b) = (1 ; 25) ; (5 ; 5) ; (25 ; 1)

\(\Rightarrow a+b=10;26\) (absurd)

Hence,there are no answers