1) Áp dụng bất đẳng Bunyakovsky dạng cộng mẫu ta có:

\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}=\frac{a^6}{abc}+\frac{b^6}{abc}+\frac{c^6}{abc}\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\)

\(=\frac{\left(a^3+b^3+c^3\right)\left(a^3+b^3+c^3\right)}{3abc}\ge\frac{3abc\left(a^3+b^3+c^3\right)}{3abc}=a^3+b^3+c^3\)

(Cauchy 3 số) Dấu "=" xảy ra khi: a = b = c

2) Áp dụng kết quả phần 1 ta có:

\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\ge\frac{\left(a^3+b^2+c^3\right)^2}{3\cdot\frac{1}{3}}=\left(a^3+b^3+c^3\right)^2\)

Dấu "=" xảy ra khi: \(a=b=c=\frac{1}{\sqrt[3]{3}}\)

11/Theo BĐT AM-GM,ta có; \(ab.\frac{1}{\left(a+c\right)+\left(b+c\right)}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)\(=\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)

Tương tự với hai BĐT kia,cộng theo vế và rút gọn ta được đpcm.

Dấu "=" xảy ra khi a= b=c

Ơ vãi,em đánh thiếu abc dưới mẫu,cô xóa giùm em bài kia ạ!

9/ \(VT=\frac{\Sigma\left(a+2\right)\left(b+2\right)}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\)

\(=\frac{ab+bc+ca+4\left(a+b+c\right)+12}{\left(ab+bc+ca\right)+4\left(a+b+c\right)+8+abc+\left(ab+bc+ca\right)}\)

\(\le\frac{ab+bc+ca+4\left(a+b+c\right)+12}{\left(ab+bc+ca\right)+4\left(a+b+c\right)+9+3\sqrt[3]{\left(abc\right)^2}}\)

\(=\frac{ab+bc+ca+4\left(a+b+c\right)+12}{ab+bc+ca+4\left(a+b+c\right)+12}=1\left(Q.E.D\right)\)

"=" <=> a = b = c = 1.

Mong là lần này không đánh thiếu (nãy tại cái tội đánh ẩu)

10/Thêm \(\frac{b}{a}-2\) ở mỗi vế ta cần chứng minh:

\(\frac{\left(a-b\right)^2}{ab}+\frac{b}{c}\ge\frac{4a}{a+c}+\frac{b}{a}-2\) (vận dùng đẳng thức \(\frac{a}{b}+\frac{b}{a}-2=\frac{a^2+b^2-2ab}{ab}=\frac{\left(a-b\right)^2}{ab}\))

\(\Leftrightarrow\frac{c\left(a-b\right)^2+ab^2}{abc}\ge\frac{4a^2+ab+bc-2a\left(a+c\right)}{a\left(a+c\right)}\)

\(\Leftrightarrow\frac{c\left(a-b\right)^2+ab^2}{abc}\ge\frac{2a^2+a\left(b-c\right)+c\left(b-a\right)}{a\left(a+c\right)}\)

\(\Leftrightarrow\frac{\left(c\left(a-b\right)^2+ab^2\right)\left(a+c\right)}{abc\left(a+c\right)}-\frac{\left(2a^2+a\left(b-c\right)+c\left(b-a\right)\right)bc}{abc\left(a+c\right)}\ge0\)

Em làm tắt tiếp:v

\(\Leftrightarrow\frac{a\left(ac^2+b^2c+ca^2+ab^2-4abc\right)}{abc\left(a+c\right)}\ge0\)\(\Leftrightarrow\frac{\left(ac^2+b^2c+ca^2+ab^2-4abc\right)}{bc\left(a+c\right)}\ge0\)

Áp dụng BĐT AM-GM ta được: \(VT\ge\frac{4\sqrt[4]{\left(abc\right)^4}-4abc}{bc\left(a+c\right)}=\frac{0}{bc\left(a+c\right)}=0\)

Ta có Q.E.D. 

P/s: Đúng không ta? Mà sao có người tk sai nhỉ?

Tham khảo: https://lazi.vn/edu/exercise/cho-a-b-c-la-cac-so-duong-thoa-man-a2-2b2-3c2-chung-minh-1-a-2-b-3-c

\(ab+bc+ac=3\)

Ta có:

\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\) ( đúng với mọi \(ab\ge1\))

Giả sử:\(ab\ge1\)

\(\Rightarrow\dfrac{2}{ab+1}+\dfrac{1}{c^2+1}\ge\dfrac{2c^2+2+ab+1}{\left(ab+1\right)\left(c^2+1\right)}=\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\)

Giả sử: \(\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\ge\dfrac{3}{2}\)(đúng)

\(\Leftrightarrow2\left(2c^2+ab+3\right)\ge3\left(ab+1\right)\left(c^2+1\right)\)

\(\Leftrightarrow4c^2+2ab+6\ge3\left(abc^2+ab+c^2+1\right)\)

\(\Leftrightarrow4c^2+2ab+6\ge3abc^2+3ab+3c^2+3\)

\(\Leftrightarrow c^2-ab-3abc^2+3\ge0\)

\(\Leftrightarrow c^2-ab-3abc^2+ab+ac+bc\ge0\)   ( vì \(ab+ac+bc=3\) )

\(\Leftrightarrow c^2+ac+bc-3abc^2\ge0\)

\(\Leftrightarrow c+a+b-3abc\ge0\)

\(\Leftrightarrow c+a+b\ge3abc\)

Ta có:

\(3\left(c+a+b\right)=\left(ab+ac+bc\right)\left(c+a+b\right)\) ( vì \(ab+ac+bc=3\) )

Áp dụng BĐT AM-GM, ta có:

\(\left(ab+ac+bc\right)\left(c+a+b\right)\ge9abc\)

\(\Rightarrow a+b+c\ge3abc\)

\(\Rightarrow\) \(\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\ge\dfrac{3}{2}\) ( luôn đúng )

\(\Rightarrow\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{3}{2}\) ( đfcm )

Dấu "=" xảy ra khi \(a=b=c=1\)

Hình như sai đề rồi bạn ạ, dấu ≥ phải là ≤

Đặt \(\left(a;b;c\right)=\left(\dfrac{y}{x};\dfrac{z}{y};\dfrac{x}{z}\right)\)

\(\Rightarrow VT=\dfrac{1}{\dfrac{y}{x}\left(\dfrac{z}{y}+1\right)}+\dfrac{1}{\dfrac{z}{y}\left(\dfrac{x}{z}+1\right)}+\dfrac{1}{\dfrac{x}{z}\left(\dfrac{y}{x}+1\right)}\)

\(VT=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\)

\(VT\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\dfrac{3\left(xy+yz+zx\right)}{2\left(xy+yz+zx\right)}=\dfrac{3}{2}\)