Question

Match the following

	Column-I		Column-II
(A)	$a,b,c>0$ then $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ can't take the value	(p)	$\frac{121}{81}$
(B)	p and q are positive real numbers and p + q = 1, then ${(p+\frac{1}{p})}^2+{(q+\frac{1}{q})}^2$ can't take the value	(q)	$\frac{343}{229}$
(C)	x, y, z are real numbers such that 0 <x, y, z< 1 and x + y + z = 2, then $\frac{xyz}{(1-x)(1-y)(1-z)}$ can take the value	(r)	$\frac{133}{15}$
(D)	x, y, z are positive real numbers such that x + y + z = 1, then \frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)} can take the value	(s)	$\frac{413}{51}$
		(t)	8

Answer 1

The least value of $\frac{a}{b+c}+\frac{b}{c+c}+\frac{c}{a+b}is\frac{3}{2}$
${(p+\frac{a}{p})}^2+{(q+\frac{1}{q})}^2=(p^2+q^2)+\frac{1}{p^2}+\frac{1}{q^2}+4$
$=(p^2+q^2)(1+\frac{1}{p^2{q}^2})+4$
We have
$\frac{1}{pq}\le 4\left(\text{Also}\right(p^1+q^1\ge \frac{1}{2})$
$\therefore$ The expression $\ge \frac{}{}(1+4^2)+4=\frac{17}{2}+4=\frac{25}{2}$
$\frac{xyz}{(1-x)(1-y)(1-z)}\ge 8$ can be shown using AM-GM in equality

Similarly, $\frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}\ge 8$ can be shown using AM-GM in equality.