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

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

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

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

Đề bài sai

Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x^2;y^2;z^2\right)\Rightarrow xyz=1\)

Đặt vế trái BĐT cần chứng minh là P, ta có:

\(P=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)

\(P=\dfrac{1}{\left(x^2+y^2\right)+\left(y^2+1\right)+2}+\dfrac{1}{\left(y^2+z^2\right)+\left(z^2+1\right)+2}+\dfrac{1}{\left(z^2+x^2\right)+\left(x^2+1\right)+2}\)

\(P\le\dfrac{1}{2xy+2y+2}+\dfrac{1}{2yz+2z+2}+\dfrac{1}{2zx+2x+2}\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{xz\left(xy+y+1\right)}+\dfrac{x}{x\left(yz+z+1\right)}+\dfrac{1}{zx+x+1}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{x.xyz+xyz+xz}+\dfrac{x}{xyz+xz+1}+\dfrac{1}{xz+x+1}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{x+1+xz}+\dfrac{x}{1+xz+1}+\dfrac{1}{xz+x+1}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)

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

Ta có \(a^2+\dfrac{1}{b+c}=a^2+\dfrac{1}{6-a}\)

Mà \(a+b+c=6\Rightarrow0\le a,b,c\le2\)

\(\Rightarrow a^2+\dfrac{1}{6-a}\ge2^2+\dfrac{1}{6-2}=\dfrac{17}{4}\)

\(\Rightarrow P=\sum\sqrt{a^2+\dfrac{1}{b+c}}=\sum\sqrt{a^2+\dfrac{1}{6-a}}\ge\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}=\dfrac{3\sqrt{17}}{2}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

Em kiểm tra lại mẫu số của biểu thức c, chắc chắn đề sai

Chia 2 vế cho \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\) BĐT trở thành:

\(\dfrac{1}{a^4\left(b+1\right)\left(c+1\right)}+\dfrac{1}{b^4\left(a+1\right)\left(c+1\right)}+\dfrac{1}{c^4\left(a+1\right)\left(b+1\right)}\ge\dfrac{3}{4}\)

Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\) \(\Rightarrow xyz=1\)

\(\dfrac{1}{a^4\left(b+1\right)\left(c+1\right)}=\dfrac{x^4}{\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)}=\dfrac{x^4yz}{\left(y+1\right)\left(z+1\right)}=\dfrac{x^3}{\left(y+1\right)\left(z+1\right)}\)

Do đó BĐT trở thành:

\(\dfrac{x^3}{\left(y+1\right)\left(z+1\right)}+\dfrac{y^3}{\left(x+1\right)\left(z+1\right)}+\dfrac{z^3}{\left(x+1\right)\left(y+1\right)}\ge\dfrac{3}{4}\)

Một bài toán quen thuộc

Đặt \(x=\sqrt{a};y=\sqrt{b};z=\sqrt{c}\) \(\Rightarrow xyz=1\)  (x;y;z > 0 do a;b;c>0)

Cần c/m : \(VT=\dfrac{y^2+z^2}{x}+\dfrac{x^2+z^2}{y}+\dfrac{x^2+y^2}{z}\ge x+y+z+3=VP\) 

Dễ dàng c/m : VT \(\ge2\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\)   (1)

Thấy : \(\dfrac{xy}{z}+\dfrac{xz}{y}\ge2x\)  . CMTT : \(\dfrac{xz}{y}+\dfrac{yz}{x}\ge2z;\dfrac{yz}{x}+\dfrac{xy}{z}\ge2y\)

Suy ra : \(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}\ge x+y+z\)

Có : \(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}\ge3\sqrt[3]{xyz}=3\)

Suy ra : \(2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\right)\ge x+y+z+3\left(2\right)\)

Từ (1) ; (2) suy ra : \(VT\ge VP\)

" = " \(\Leftrightarrow x=y=z=1\Leftrightarrow a=b=c=1\)

*** $a,b,c>0$ thôi chứ không lớn hơn $1$ bạn nhé. $a,b,c>1$ thì $abc>1$ mất rồi.

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Vì $a, b, c>0$ thỏa mãn $abc=1$ nên tồn tại $x,y,z>0$ sao cho:

$(a,b,c)=(\frac{x^2}{yz}, \frac{y^2}{xz}, \frac{z^2}{xy})$
Khi đó, áp dụng BĐT Cauchy_Schwarz:
$P=\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}$

$\geq \frac{(x+y+z)^2}{x^2+2yz+y^2+2xz+z^2+2xy}=\frac{(x+y+z)^2}{(x+y+z)^2}=1$

Vậy $P_{\min}=1$ khi $x=y=z\Leftrightarrow a=b=c=1$

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?