Condition on a,b and c satisfying an equation(TIFR GS 2017)

Let $a,b,c$ be positive real numbers satisfying $$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=16,$$ then $a+b+c=3$.

I thought about the application of the AM-GM-HM inequality, but in vain. I also thought about splitting 16 into factors and comparing but went nowhere. Any ideas. Thanks beforehand.

Hint. Use Cauchy–Schwarz inequality with vectors $${\bf u}=(1,\sqrt{a},\sqrt{b},\sqrt{c})\quad\mbox{and}\quad{\bf v}=(1,1/\sqrt{a},1/\sqrt{b},1/\sqrt{c}).$$

When does equality hold?

by AM-GM we have $$\frac{1+a+b+c}{4}\geq \sqrt[4]{abc}$$ and by AM-GM again we get $$\frac{1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{4}\geq \frac{1}{\sqrt[4]{abc}}$$ multiplying both we obtain $$16\geq 16$$ thus we have $$a=b=c=1$$

By the AM-HM inequality $$\frac{1 + a + b + c}{4} \geq \frac{4}{\frac{1}{1} + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$ and equality holds if and only if $1 = a = b = c$. Since the problem statement tells us that equality holds we find $a + b + c = 3$.