Cdtrjyty

Let $a,b,c$ be positive real number such that
$$(1+a+b+c)left(1+frac{1}{a}+frac{1}{b}+frac{1}{c}right)=16$$
then $a+b+c=3$

I think this is trivial but I don't know how to prove this fact.

Any Help will be appreciated

Use the AM-HM inequality of $1, a, b, c$.

So $$frac{1 + a + b + c}{4} geq frac{4}{1 + frac{1}{a} + frac{1}{b} + frac{1}{c}}$$

This gives $$(1 + a + b + c)bigg(1 + frac{1}{a} + frac{1}{b} + frac{1}{c}bigg) geq 16$$.

The equality becomes an equation only when all four elements are equal.

i.e. $1=a=b=c$. So $a + b + c = 3$.

Using Cauchy-Schwarz inequality you can even show that $a=b=c=1$:
$$(1+a+b+c)left(1+frac{1}{a}+frac{1}{b}+frac{1}{c}right) = left(1^2 + sum_{cyc}left(sqrt{a}right)^2 right)left(1^2 + sum_{cyc}left(frac{1}{sqrt{a}}right)^2 right)$$ $$stackrel{C.-S.}{geq}(1+1+1+1)^2 = 16$$
Equality holds for $(1: sqrt{a} : sqrt{b} : sqrt{c})^T = lambda cdot left( 1: frac{1}{sqrt{a}} : : frac{1}{sqrt{b}} : : frac{1}{sqrt{c}}right)^T Rightarrow a=b=c = 1$

Hint:

Using AM HM inequality for non-negative numbers,

$$dfrac{sum_{r=1}^na_r}ngedfrac n{sum_{r=1}^ndfrac1{a_r}}$$