Question

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

Collected in the board: Inequality

Steven Zheng posted 1 year ago

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

\therefore \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


Steven Zheng posted 4 months ago

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