Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq (1+1+1)^2\)

\(\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 9\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\)

áp dụng BĐT:
1/a +1/b+1/c>= 9/a+b+c mà a+b+c=1

=>

a,b,c là các số dương nên \(\left(a+b+c\right)>=3\cdot\sqrt[3]{abc}\)

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=3\cdot\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}\)

Do đó: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=3\cdot\sqrt[3]{abc}\cdot3\cdot\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\cdot\sqrt[3]{a\cdot b\cdot c\cdot\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\)

Áp dụng bất đẳng thức Cauchy-Schwarz:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{1}=9\)

Dấu "=" xảy ra khi: \(a=b=c=\dfrac{1}{3}\)

Áp dụng BĐT Svacxo ta được

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{\left(a+b+c\right)}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)

Vậy BĐT được chứng minh

Cauchy-Schwarz: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{1}=9\)

Cách khác:

Đặt \(A=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\)

\(A=\left(1+\dfrac{a+b}{a}\right)\left(1+\dfrac{a+b}{b}\right)\)

\(A=\left(2+\dfrac{b}{a}\right)\left(2+\dfrac{a}{b}\right)\)

\(A=4+2\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+1\)

\(A\ge4+2\cdot2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}+1=9\left(AM-GM\right)\left(đpcm\right)\)

( 1 + \(\dfrac{1}{a}\))\(\left(1+\dfrac{1}{b}\right)\) ≥ 9

Biến đổi VT Ta có : VT = \(\dfrac{a+1}{a}.\dfrac{b+1}{b}\)

= \(\dfrac{2a+b}{a}.\dfrac{2b+a}{b}\)

=\(\left(2+\dfrac{b}{a}\right)\left(2+\dfrac{a}{b}\right)\)

= 4 + \(\dfrac{2a}{b}+\dfrac{2b}{a}+\dfrac{b}{a}.\dfrac{a}{b}\)

= 5 + 2( \(\dfrac{a}{b}+\dfrac{b}{a}\) ) ( *)

Áp dụng BĐT : \(\dfrac{x}{y}+\dfrac{y}{x}\) ≥ 2( x > 0 ; y > 0) ( ** )

Từ ( * ; **) ⇒ 5 + 2( \(\dfrac{a}{b}+\dfrac{b}{a}\) ) ≥ 5 + 4 = 9 ( đpcm )

Lời giải +HD chi tiết

\(A=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(A=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\) {vì (a+b+c=1}

\(A=\left(\dfrac{a+b+c}{a}\right)+\left(\dfrac{a+b+c}{b}\right)+\left(\dfrac{a+b+c}{c}\right)\) {nhân pp}
\(A=\left(\dfrac{a}{a}+\dfrac{b}{a}+\dfrac{c}{a}\right)+\left(\dfrac{a}{b}+\dfrac{b}{b}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{c}{c}\right)\){tách nhỏ ra}

\(A=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\) ghép lại theo định hướng

\(\left\{{}\begin{matrix}\dfrac{a}{b}=x\\\dfrac{b}{c}=y\\\dfrac{a}{c}=z\end{matrix}\right.\) \(\Rightarrow A=3+\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)+\left(z+\dfrac{1}{z}\right)\) {đổi biến viêt cho gọn }

\(A=3+2.3+\left(\sqrt{x}-2+\sqrt{\dfrac{1}{x}}\right)+\left(\sqrt{y}-2+\sqrt{\dfrac{1}{y}}\right)+\left(\sqrt{z}-2+\sqrt{\dfrac{1}{z}}\right)\)

{định hướng ghép bp}

\(A=9+\left(\sqrt{x}-\sqrt{\dfrac{1}{x}}\right)^2+\left(\sqrt{y}-\sqrt{\dfrac{1}{y}}\right)^2+\left(\sqrt{z}-\sqrt{\dfrac{1}{z}}\right)^2\)

\(\sum\left(x-\dfrac{1}{x}\right)^2\ge0\Rightarrow9+\sum\left(x-\dfrac{1}{x}\right)^2\ge9\Rightarrow A\ge9\)Kết thúc

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)

Áp dụng BĐT cauchy-schwarz dạng engel ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{a+b+c}=9\)

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

=> (a+b).\(\left(\dfrac{1}{b}+\dfrac{1}{b}\right)\ge\left(a+b\right).\dfrac{4}{a+b}=4\left(dpcm\right)\)

b)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+b+c}\)

=>\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(a+b+c\right).\dfrac{9}{a+b+c}=9\left(dpcm\right)\)