Question

If a, b, c are real numbers such that a + 2b + c = 4 then max. value of ab + bc + ca is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is B 4

Given $(a+b)+(b+c)=4$
But ${\left[\frac{(a+b)+(b+c)}{2}\right]}^2\ge \left[\right(a+b\left)\right(b+c\left)\right],A.M\ge G.M.$
or ${\left(\frac{4}{2}\right)}^2\ge ab+bc+ca+b^2$
$4-b^2\ge \sum ab$ or $\sum ab\le 4-b^2$
$\therefore$ Max. value of $\sum ab=4\Rightarrow \left(b\right)$