\(VT=\dfrac{\left(a+c\right)^2}{\left(a+c\right)\left(a+b\right)}+\dfrac{\left(b+d\right)^2}{\left(b+c\right)\left(b+d\right)}+\dfrac{\left(c+a\right)^2}{\left(c+a\right)\left(c+d\right)}+\dfrac{\left(d+b\right)^2}{\left(d+a\right)\left(d+b\right)}\)

\(VT\ge\dfrac{\left(2a+2b+2c+2d\right)^2}{\left(a+b\right)\left(a+c\right)+\left(b+c\right)\left(b+d\right)+\left(a+c\right)\left(c+d\right)+\left(a+d\right)\left(b+d\right)}=\dfrac{4\left(a+b+c+d\right)^2}{\left(a+b+c+d\right)^2}=4\)

Dấu "=" xảy ra khi \(a=b=c=d\)

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{b}+\frac{1}{c}\geq \frac{4}{b+c}\)

\(\Rightarrow \frac{a}{b}+\frac{a}{c}\geq \frac{4a}{b+c}(1)\)

Hoàn toàn tương tự: \(\frac{b}{c}+\frac{b}{a}\geq \frac{4b}{c+a}(2)\)

\(\frac{c}{a}+\frac{c}{b}\geq \frac{4c}{a+b}(3)\)

Lấy \((1)+(2)+(3)\Rightarrow \frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}\geq 4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)

\(\Leftrightarrow \frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\geq 4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)

Ta có đpcm

Dấu bằng xảy ra khi $a=b=c$

\(\left(a+b+c\right)\left(\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(c+a\right)^2}+\dfrac{c}{\left(a+b\right)^2}\right)\ge\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\ge\dfrac{9}{4}\)

\(\Rightarrow\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(c+a\right)^2}+\dfrac{c}{\left(a+b\right)^2}\ge\dfrac{9}{4\left(a+b+c\right)}\)

Dấu "=" xảy ra khi \(a=b=c\)

Bài này có bạn giải rồi:

Cho các số thực dương a,b,c.Chứng minh rằng :\(\dfrac{b\left(2a-b\right)}{a\left(b+c\right)}+\dfrac{c\left(2b-c\right)}{... - Hoc24

\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge3+\dfrac{2a^2+2b^2+2c^2-2\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)

\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge5-\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)

\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)

Do \(\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}=\dfrac{2a^2}{ab+ac}+\dfrac{2b^2}{bc+ab}+\dfrac{2c^2}{ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}\)

Nên ta chỉ cần chứng minh:

\(\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)

Điều này hiển nhiên đúng do:

\(VT=\dfrac{2}{3}.\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}+\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\)

\(VT\ge2\sqrt{\dfrac{12\left(a+b+c\right)^2\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)\left(a+b+c\right)^2}}+\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=5\)

Dấu "=" xảy ra khi \(a=b=c\)

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

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

a) sai đề

b) để ý rằng :Theo AM-GM

\(VT=\dfrac{a+b}{2\sqrt[3]{abc}}+\dfrac{b+c}{2\sqrt[3]{abc}}+\dfrac{c+a}{2\sqrt[3]{abc}}+\dfrac{8abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge4\)

Dấu = xảy ra khi a=b=c.

P/s: Min ra xấp xỉ \(14,4809\)( wolframalpha.com)

`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`

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 \(a=b=c\)