<s>[INST] 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.
 [/INST] 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 $a + b + c = 3$.
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