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This question already has an answer here:

Let $a$, $b$ and $c$ be positive real numbers such that $$(1+a+b+c)\big(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)=16.$$ Prove that $$a+b+c=3.$$

I am not able to get how to prove that? Thanks for help in advance .

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marked as duplicate by Martin R, Jack D'Aurizio inequality Sep 23 '17 at 13:44

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By arithmetic-harmonic mean inequality, $$\frac{1+a+b+c}{4} \ge \frac{4}{\frac{1}{1}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}.$$ Consequently, $$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \ge 16.$$ The equality in the inequality holds when all terms are equal. That is, $1=a=b=c$. Therefore, $1+a+b+c=4$.

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By the Cauchy-Schwarz inequality on the two vectors $$ \left(1, \sqrt a, \sqrt b, \sqrt c\right) \text{ and }\left(1, \frac1{\sqrt a}, \frac1{\sqrt b}, \frac1{\sqrt c}\right) $$ we get $$ \left(\left(1, \sqrt a, \sqrt b, \sqrt c\right)\cdot \left(1, \frac1{\sqrt a}, \frac1{\sqrt b}, \frac1{\sqrt c}\right)\right)^2\leq (1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = 16 $$ and equality in the Cauchy-Schwarz inequality implies that the two vectors are parallel.

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By C-S $$(1+a+b+c)\big(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)\geq(1+1+1+1)^2=16$$ The equality occurs for $$(1,\sqrt{a},\sqrt{b},\sqrt{c})||(1,\frac{1}{\sqrt{a}},\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c}}\big)$$ or for $a=b=c=1$, which gives $a+b+c=3$ and we are done!

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Cauchy-schwarz inequality state that :

  • For any $x_1, y_1, z_1, w_1; x_2, y_2, z_2, w_2 \ \in \mathbb{R}^+$; we have: $$ \left(\sqrt{x_1 \cdot x_2} + \sqrt{y_1 \cdot y_2} + \sqrt{z_1 \cdot z_2} + \sqrt{w_1 \cdot w_2} \right) ^ 2 \leq \\ \left(x_1 + y_1 + z_1 + w_1\right) \cdot \left(x_2 + y_2 + z_2 + w_2\right) \ . $$


So we can conclude that:

$$ \begin{align} & \left( \sqrt{\frac{1}{1}} + \sqrt{\frac{a}{a}} + \sqrt{\frac{b}{b}} + \sqrt{\frac{c}{c}} \right) ^ 2 & \leq \ \ \ \ & \left(1 + \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) & \cdot \ \ \ \ \ \ \ \ & (1+a+b+c) & \Longrightarrow \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( 4 \right) ^ 2 & \leq \ \ \ \ & \left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) & \cdot \ \ \ \ \ \ \ \ & \ \ \ \ \ \ \ \ (1+a+b+c) & \Longrightarrow \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 16 & \leq \ \ \ \ & \left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) & \cdot \ \ \ \ \ \ \ \ & \ \ \ \ \ \ \ \ (1+a+b+c) & \Longrightarrow \end{align} $$


Note that equlality holds if and only if $a=1, b=1, c=1$;
for which one can see the value of $\left(a+b+c\right)$ is equal to $3$.

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