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

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

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

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

do \(a,b,c\ge1\)\(=>\left\{{}\begin{matrix}b+c\ge2\\c+a\ge2\\a+b\ge2\end{matrix}\right.\)

\(=>\left\{{}\begin{matrix}a\left(b+c\right)\ge2a\\b\left(c+a\right)\ge2b\\c\left(a+b\right)\ge2c\end{matrix}\right.\)

\(=>\) biểu thức đề bài cho\(\ge2\left(a+b+c+\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\)

\(2\left(1+1+1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\right)=9\)

dấu= xảy ra<=>a=b=c=1

\(a;b;c\ge1\Rightarrow\left\{{}\begin{matrix}ab;bc;ca\ge1\\ab+bc+ca\ge3\end{matrix}\right.\)

Ta có:

\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}=\dfrac{a^2+b^2+2}{a^2b^2+a^2+b^2+1}=1-\dfrac{a^2b^2-1}{a^2b^2+a^2+b^2+1}\ge1-\dfrac{a^2b^2-1}{a^2b^2+2ab+1}=\dfrac{2}{ab+1}\)

Tương tự: \(\dfrac{1}{a^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{2}{ac+1}\) ; \(\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{2}{bc+1}\)

Cộng vế: \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{1}{ab+1}+\dfrac{1}{bc+1}+\dfrac{1}{ca+1}\)

Do đó: \(VT\ge2\left(ab+bc+ca+\dfrac{1}{ab+1}+\dfrac{1}{bc+1}+\dfrac{1}{ca+1}\right)\)

\(VT\ge2\left(ab+bc+ca+\dfrac{9}{ab+bc+ca+3}\right)\)

Đặt \(ab+bc+ca=x\ge3\Rightarrow VT\ge2\left(x+\dfrac{9}{x+3}\right)\)

\(VT\ge2\left(\dfrac{x+3}{4}+\dfrac{9}{x+3}+\dfrac{3x}{4}-\dfrac{3}{4}\right)\ge2\left(2\sqrt{\dfrac{9\left(x+3\right)}{4\left(x+3\right)}}+\dfrac{3}{4}.3-\dfrac{3}{4}\right)=9\)

wow

wow, chắc xu học lớp 9

\(\dfrac{a^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{a^3\left(a+b\right)\left(b+c\right)}{64}}=\dfrac{3a}{4}\)

Tương tự:

\(\dfrac{b^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{c+a}{8}\ge\dfrac{3b}{4}\)

\(\dfrac{c^3}{\left(c+a\right)\left(a+b\right)}+\dfrac{c+a}{8}+\dfrac{a+b}{8}\ge\dfrac{3c}{4}\)

Cộng vế:

\(VT+\dfrac{4\left(a+b+c\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)

\(\Rightarrow VT\ge\dfrac{a+b+c}{4}\)

Dấu "=" xảy ra khi \(a=b=c\)

gợi ý đy :

thèn này ko làm mà lôi BTVN ra hỏi lmj z ?

@@@ @@@

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

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

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

\(=\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+3\)

Áp dụng BĐT Cô - si cho 2 số không âm:

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)

\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2\)

\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}=2\)

Suy ra:

\(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}+3\ge2+2+2+3=9\)

Dấu "=" xảy ra khi: a = b = c

Áp dụng BĐT Cauchy dạng Engel , ta có :

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

⇔ \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{≥}\left(a+b+c\right).\dfrac{9}{a+b+c}\)

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

\("="\text{⇔}a=b=c\)

Áp dụng bất đẳng thức Cô - si cho 3 số không âm ta có :

\(a+b+c\ge3\sqrt[3]{abc}=3\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}=3\)

Nhân vế theo vế ta có :

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)

Dấu " = " xảy ra khi a = b = c .

`sqrta+sqrtb+sqrtc=2`

`<=>(sqrta+sqrtb+sqrtc)^2=4`

`<=>a+b+c+2sqrt{ab}+2sqrt{bc}+2sqrt{ca}=4`

`<=>2sqrt{ab}+2sqrt{bc}+2sqrt{ca}=4-(a+b+c)=4-2-2`

`<=>sqrt{ab}+sqrt{bc}+sqrt{ca}=1`

`=>a+1=a+sqrt{ab}+sqrt{bc}+sqrt{ca}=sqrta(sqrta+sqrtb)+sqrtc(sqrta+sqrtb)=(sqrta+sqrtb)(sqrta+sqrtc)`

Tương tự:`b+1=(sqrtb+sqrta)(sqrtb+sqrtc)`

`c+1=(sqrtc+sqrta)(sqrtc+sqrtb)`

`=>VT=sqrta/((sqrta+sqrtb)(sqrta+sqrtc))+sqrtb/((sqrtb+sqrta)(sqrtb+sqrtc))+sqrtc/((sqrtc+sqrta)(sqrtc+sqrtb))`

`=>VT=(sqrta(sqrtb+sqrtc)+sqrtb(sqrtc+sqrta)+sqrtc(sqrta+sqrtb))/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=(sqrt{ab}+sqrt{ac}+sqrt{bc}+sqrt{ab}+sqrt{ac}+sqrt{bc})/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=(2(sqrt{ab}+sqrt{bc}+sqrt{ca}))/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=2/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=2/\sqrt{[(sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta)]^2}`

`=2/\sqrt{(sqrta+sqrtb)(sqrta+sqrtc)(sqrtb+sqrta)(sqrtb+sqrtc)(sqrtc+sqrta)(sqrtc+sqrtb)}`

`=2/\sqrt{(1+a)(1+b)(1+c)}=>đpcm`

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

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

= \(3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)

\(\ge3+2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}+2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}\)

\(\ge3+2+2+2=9\left(đpcm\right)\)

vì a,b,c là các số dương nên ta có:

\(a+b+c\ge3\sqrt[3]{abc}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)

nhân hai vế vs nhau, ta có

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)