For a given condition is it true that $a+b+c=3$.

Question

Suppose a, b, c are positive real numbers such that

$$(1+a+b+c)\left(1+\frac 1a+\frac 1b+\frac 1c\right)=16$$ Then is it true that we must have $a+b+c=3$ ?

Please help me to solve this. Thanks in advance.

Answer 1

By Cauchy-Schwarz $$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq\left(1\cdot1+\sqrt{a}\cdot\frac{1}{\sqrt{a}}+\sqrt{b}\cdot\frac{1}{\sqrt{b}}+\sqrt{c}\cdot\frac{1}{\sqrt{c}}\right)^2= 16$$ The equality occurs for $a=b=c=1$, which says that $a+b+c=3$.

Answer 2

Remember the inequality between the arithmetic and the harmonic mean $$\frac{1+a+b+c}{4}\geq \frac{4}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1}$$ for all $$a,b,c>0$$ the equality holds if $$a=b=c=1$$ thus $$a+b+c=3$$

Answer 3

Expand your product to get $$4+\left(a+\frac 1a\right)+\left(b+\frac 1b\right)+\left(c+\frac 1c\right)+\left(\frac ab+\frac ba\right)+\left(\frac ac+\frac ca\right)+\left(\frac bc+\frac cb\right)=16$$

Now, the arithmetic-geometric inequality tells us that (for $x>0$): $$\left(x+\frac 1x\right)≥2$$ and that equality only holds when $x=1$. This quickly implies that each of the variable terms in the expanded expression must be $2$ and that $a=b=c=1$ and we are done.