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

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

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

Áp dụng BĐT cô si cho 2 số ta có

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

⇔\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)

⇔\(2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge4\)

⇔ A ≥4

=> Min A =4

dấu "=" xảy ra khi

\(\dfrac{a}{b}=\dfrac{b}{a}\)

⇔a²=b²

⇔a=b

vậy Min A =4 khi a=b

\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)=8\)

\(\Leftrightarrow\dfrac{a+b}{a}\times\dfrac{b+c}{b}\times\dfrac{a+c}{c}=8\)

\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=8abc\)

~*~*~*~*~

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

\(=\dfrac{3}{4}+\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{ac}{\left(c+a\right)\left(a+b\right)}\) (1)

\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{b}{b+c}-\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{c}{c+a}-\dfrac{ac}{\left(c+a\right)\left(a+b\right)}\)

\(=\dfrac{3}{4}\)

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

\(=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{a}{a+b}\times\dfrac{c}{b+c}+\dfrac{b}{b+c}\times\dfrac{a}{a+c}+\dfrac{c}{a+c}\times\dfrac{b}{a+b}\)

\(=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)}{\left(a+c\right)\left(b+c\right)\left(a+b\right)}=\dfrac{3}{4}\)

\(\Leftrightarrow ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)=\dfrac{3}{4}\times8abc\)

\(\Leftrightarrow ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)+2abc=8abc\)

\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=8abc\) luôn đúng

=> (1) đúng

Bạn cũng có thể giải bằng cách đặt \(x=\dfrac{a}{a+b};y=\dfrac{b}{b+c};z=\dfrac{c}{a+c}\).

1.

BĐT cần chứng minh tương đương:

\(\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\)

Ta có:

\(\left(ab-1\right)^2=a^2b^2-2ab+1=a^2b^2-a^2-b^2+1+a^2+b^2-2ab\)

\(=\left(a^2-1\right)\left(b^2-1\right)+\left(a-b\right)^2\ge\left(a^2-1\right)\left(b^2-1\right)\)

Tương tự: \(\left(bc-1\right)^2\ge\left(b^2-1\right)\left(c^2-1\right)\)

\(\left(ca-1\right)^2\ge\left(c^2-1\right)\left(a^2-1\right)\)

Do \(a;b;c\ge1\)  nên 2 vế của các BĐT trên đều không âm, nhân vế với vế:

\(\left[\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\right]^2\ge\left[\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\right]^2\)

\(\Rightarrow\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Câu 2 em kiểm tra lại đề có chính xác chưa

2.

Câu 2 đề thế này cũng làm được nhưng khá xấu, mình nghĩ là không thể chứng minh bằng Cauchy-Schwaz được, phải chứng minh bằng SOS

Không mất tính tổng quát, giả sử \(c=max\left\{a;b;c\right\}\)

\(\Rightarrow\left(c-a\right)\left(c-b\right)\ge0\) (1)

BĐT cần chứng minh tương đương:

\(\dfrac{1}{a}-\dfrac{a+b}{bc+a^2}+\dfrac{1}{b}-\dfrac{b+c}{ac+b^2}+\dfrac{1}{c}-\dfrac{c+a}{ab+c^2}\ge0\)

\(\Leftrightarrow\dfrac{b\left(c-a\right)}{a^3+abc}+\dfrac{c\left(a-b\right)}{b^3+abc}+\dfrac{a\left(b-c\right)}{c^3+abc}\ge0\)

\(\Leftrightarrow\dfrac{c\left(b-a\right)+a\left(c-b\right)}{a^3+abc}+\dfrac{c\left(a-b\right)}{b^3+abc}+\dfrac{a\left(b-c\right)}{c^3+abc}\ge0\)

\(\Leftrightarrow c\left(b-a\right)\left(\dfrac{1}{a^3+abc}-\dfrac{1}{b^3+abc}\right)+a\left(c-b\right)\left(\dfrac{1}{a^3+abc}-\dfrac{1}{c^3+abc}\right)\ge0\)

\(\Leftrightarrow\dfrac{c\left(b-a\right)\left(b^3-a^3\right)}{\left(a^3+abc\right)\left(b^3+abc\right)}+\dfrac{a\left(c-b\right)\left(c^3-a^3\right)}{\left(a^3+abc\right)\left(c^3+abc\right)}\ge0\)

\(\Leftrightarrow\dfrac{c\left(b-a\right)^2\left(a^2+ab+b^2\right)}{\left(a^3+abc\right)\left(b^3+abc\right)}+\dfrac{a\left(c-b\right)\left(c-a\right)\left(a^2+ac+c^2\right)}{\left(a^3+abc\right)\left(c^3+abc\right)}\ge0\)

Đúng theo (1)

Dấu "=" xảy ra khi \(a=b=c\)

Với \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}b+c=-a\\c+a=-b\\a+b=-c\end{matrix}\right.\)

\(B=\dfrac{a+b}{a}\cdot\dfrac{a+c}{c}\cdot\dfrac{b+c}{b}=\dfrac{-abc}{abc}=-1\)

Với \(a+b+c\ne0\)

\(\dfrac{a+b-2021c}{c}=\dfrac{b+c-2021a}{a}=\dfrac{c+a-2021b}{b}=\dfrac{-2019\left(a+b+c\right)}{a+b+c}=-2019\\ \Leftrightarrow\left\{{}\begin{matrix}a+b-2021c=-2019c\\b+c-2021a=-2019a\\c+a-2021b=-2019b\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)

\(B=\dfrac{a+b}{a}\cdot\dfrac{a+c}{c}\cdot\dfrac{b+c}{b}=\dfrac{2a\cdot2b\cdot2c}{abc}=8\)

Với a+b+c=0⇔⎧⎪⎨⎪⎩b+c=−ac+a=−ba+b=−ca+b+c=0⇔{b+c=−ac+a=−ba+b=−c

Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.

may cai nay tuong hoi truoc co nguoi dang roi ma

ta có:

\(\sqrt{\dfrac{\left(a+b\right).\left(a+c\right)}{a^2}}\le\dfrac{1}{2}.\left(\dfrac{a+b}{a}+\dfrac{a+c}{a}\right)=a+\dfrac{b}{2}+\dfrac{c}{2}\)

tương tự thì ta có:

\(VP\le3+2\left(a+b+c\right)\)

\(VP=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=3+\dfrac{2}{ab}+\dfrac{2}{ac}+\dfrac{2}{bc}\)

từ các điều trên ta thấy cần CM:

\(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge a+b+c\)

bạn tự CM nốt ạ

Lời giải:

Áp dụng BĐT AM-GM cho các số dương ta có:

\(\frac{a^3}{(a+1)(b+1)}+\frac{a+1}{8}+\frac{b+1}{8}\geq 3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)

\(\frac{b^3}{(b+1)(c+1)}+\frac{b+1}{8}+\frac{c+1}{8}\geq 3\sqrt[3]{\frac{b^3}{64}}=\frac{3b}{4}\)

\(\frac{c^3}{(c+1)(a+1)}+\frac{c+1}{8}+\frac{a+1}{8}\geq 3\sqrt[3]{\frac{c^3}{64}}=\frac{3c}{4}\)

Cộng theo vế:

\(\Rightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}+\frac{a+b+c+3}{4}\geq \frac{3}{4}(a+b+c)\)

\(\Leftrightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}+\frac{3}{2}\geq \frac{9}{4}\)

\(\Leftrightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}\geq \frac{3}{4}\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c=1\)

a) Áp dụng BĐT AM - GM:

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

Dấu '=' xảy ra <=> a=b=1

b) Cũng áp dụng BĐT AM- GM nhưng cho 3 số

c) Áp dụng BĐT AM- GM a+b>= 2\(\sqrt{ab}\)

\(\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\) >= 8\(\sqrt{ab.bc.ca}\) = 8abc

Dấu '=' xảy ra <=> a=b=c