[jacks]

If  \( a,b,c>0 \) and \(  a+b+c=1 \). Then maximum value of \( (1-a)(2-b)(3-c) \)

[Luis Fuentes]

Hi

If  \( a,b,c>0 \) and \(  a+b+c=1 \). Then maximum value of \( (1-a)(2-b)(3-c) \)

Consider:

\( f(a,b)=(1-a)(2-b)(3-(1-a-b))=(1-a)(2-b)(2+a+b) \) with \( a,b\geq 0 \) \( a+b\leq 1 \).

We have:

\( \dfrac{\partial f}{\partial a}(a,b)=-(2-b)(2+a+b)+(1-a)(2-b)=(2-b)(-1-2a-b)<0 \)

So the function is decresing on \( a \):

\( f(a,b)<f(0,b)=4-b^2<4=f(0,0) \)

The maximum \( 4 \) is reached when \( (a,b,c)=(0,0,1) \).

Under the hypothesis \( a,b,c>0 \) the maximum does not exist and by continuity, the function takes values ​​as close to \( 4 \) as we want.

Best regards.

[thadeu]

If  \( a,b,c>0 \) and \(  a+b+c=1 \). Then maximum value of \( (1-a)(2-b)(3-c) \)

Por $$MA\geq{MG}$$
$$(2-b)(3-c)\leq{(\displaystyle\frac{2-b+3-c}{2}})^2=(\displaystyle\frac{5-(1-a)}{2})^2=(\displaystyle\frac{4+a}{2})^2$$
Asi tenemos que
$$(1-a)(2-b)(3-c)\leq{(1-a)(\displaystyle\frac{4+a}{2})^2}$$
Es fácil ver que
$$f(a)=(1-a)(\displaystyle\frac{4+a}{2})^2$$
Es decreciente  en $$[0,1]$$......
Y la conclusión sigue

[jacks]

Thanks Admin  and Thanks Thadeu