Solution for 16

Consider,
(a+b+c)² = a²+b²+c²+2(ab+bc+ca)
 We know that, square of any real number is greater than or equal to zero
So, 
(a+b+c)² ≥ 0
⇒ a²+b²+c²+2(ab+bc+ca)≥ 0
⇒ 14+2(ab+bc+ca)≥ 0
⇒ 2(ab+bc+ca)≥ 0-14
⇒ (ab+bc+ca)≥ -14/2
⇒ (ab+bc+ca)≥ -7

So, ans is option 4