Question

Find the correct inequality if $a+b>0$ and $c+d>0$

• A. $a+b+c>0$
• B. $ac+bd>0$
• C. $a^2+b^2>0$
• D. $d(a+b)>0$
• E. $a+b>c+d$

Answer 1

Answer: (C)

$a^2+b^2>0$

Given $a+b>0$ and $c+d>0$ then correct option is

(A) : $a+b+c>0$ is wrong because we can no t comment od value of $c$, it is wrong if $a=1,b=1$ and $c=-5,d=6$,

$c+d=1>0$, $a+b+c<0$

(B) : $ac+bd>0$ is wrong if $a=3,b=1,c=-5,d=6$ even in this case above given condition are true

(C) : $a^2+b^2>0$ is correct, because square of any number is always positive

(D) : $d(a+b)>0$ is wrong, if $c>d$ and $d$ is negative and $a,b$ are positive

(E) : $a+b>c+d$ is wrong because we can not comment on which number is big