If $$ a,b,c∈(0, ∞) $$ and $$ a+b+c=1 $$ , prove $$ 1/a+1/b+1/c≥9 $$

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Uniteasy Admin · Accepted Answer

$$(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} )  $$
$$=( \dfrac{a}{b}+\dfrac{b}{a})+(\dfrac{a}{c}+\dfrac{c}{a})+(\dfrac{b}{c}+\dfrac{c}{b})+3      $$
$$\because a,b,c >0 $$
$$(\dfrac{\sqrt{a} }{\sqrt{b} } -\dfrac{\sqrt{b} }{\sqrt{a} } )^2\geq 0 $$
$$	herefore  \dfrac{a}{b}+\dfrac{b}{a} \geq 2\dfrac{\sqrt{a} }{\sqrt{b} }  \cdotp \dfrac{\sqrt{b} }{\sqrt{a} }  =2$$
Similarly,
we can get
$$\dfrac{a}{c}+\dfrac{c}{b}\geq 2$$
$$\dfrac{b}{c}+\dfrac{c}{b}\geq 2$$
 Therefore, 
$$(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} ) \geq 2+2+2+3=9  $$
Substitute $$a+b+c=1$$, we get,
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 9 $$

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