Giả thiết thiếu rồi em, chỗ \(\dfrac{1}{x+1}+...\) thiếu đoạn sau nữa

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

Ta cần chứng minh: \(ab+bc+ca\le\dfrac{3}{2}\)

Thật vậy, ta có:

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

\(\Rightarrow a^2+b^2+c^2+3\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Rightarrow ab+bc+ca\le\dfrac{3}{2}\) (đpcm)

Ta có:\(\frac{4+4\sqrt{1+x^2}}{4x}\le\frac{4+5+x^2}{4x}=\)\(\frac{x^2+9}{4x}\)Tương tự ta đc P\(\le\frac{x+y+z}{4}+\frac{9}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=\frac{1}{4}\left(x+y+z\right)+\frac{9}{4}\left(\frac{xy+yz+zx}{xyz}\right)\)\(\le\frac{1}{4}\left(x+y+z\right)+\frac{9}{4}\cdot\frac{\left(x+y+z\right)^2}{3\left(x+y+z\right)}\)\(=x+y+z\)

Dấu '='xảy ra <=>\(\hept{\begin{cases}x+y+z=xyz\\x=y=z\end{cases}\Rightarrow x=y=z=}\)\(\frac{1}{\sqrt{3}}\)

1.

Áp dụng BĐT Cauchy-Schwarz:

\(\dfrac{a}{2a+a+b+c}=\dfrac{a}{25}.\dfrac{\left(2+3\right)^2}{2a+a+b+c}\le\dfrac{a}{25}\left(\dfrac{2^2}{2a}+\dfrac{3^2}{a+b+c}\right)=\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{a}{a+b+c}\)

Tương tự:

\(\dfrac{b}{3b+a+c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{b}{a+b+c}\)

\(\dfrac{c}{a+b+3c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{c}{a+b+c}\)

Cộng vế:

\(VT\le\dfrac{6}{25}+\dfrac{9}{25}.\dfrac{a+b+c}{a+b+c}=\dfrac{3}{5}\)

Dấu "=" xảy ra khi \(a=b=c\)

2.

Đặt \(\dfrac{x}{x-1}=a;\dfrac{y}{y-1}=b;\dfrac{z}{z-1}=c\)

Ta có: \(\dfrac{x}{x-1}=a\Rightarrow x=ax-a\Rightarrow a=x\left(a-1\right)\Rightarrow x=\dfrac{a}{a-1}\)

Tương tự ta có: \(y=\dfrac{b}{b-1}\) ; \(z=\dfrac{c}{c-1}\)

Biến đổi giả thiết:

\(xyz=1\Rightarrow\dfrac{abc}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=1\)

\(\Rightarrow abc=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)

\(\Rightarrow ab+bc+ca=a+b+c-1\)

BĐT cần chứng minh trở thành:

\(a^2+b^2+c^2\ge1\)

\(\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\ge1\)

\(\Leftrightarrow\left(a+b+c\right)^2-2\left(a+b+c-1\right)\ge1\)

\(\Leftrightarrow\left(a+b+c-1\right)^2\ge0\) (luôn đúng)

Bài 1:

Ta có: \(\dfrac{2a}{\sqrt{1+a^2}}=\dfrac{2a}{\sqrt{ab+bc+ca+a^2}}=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)

\(\dfrac{b}{\sqrt{1+b^2}}=\dfrac{b}{\sqrt{ab+bc+ca+b^2}}=\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}\)

\(\dfrac{c}{\sqrt{1+c^2}}=\dfrac{c}{\sqrt{ab+bc+ca+c^2}}=\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)

Vậy \(P=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)

Áp dụng BĐT AM-GM ta có:

\(P\le a\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+b\left(\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{a+c}\right)+c\left(\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{a+c}\right)=\dfrac{9}{4}\)

Bài 2:

Ta có:

\(\dfrac{1+\sqrt{1+x^2}}{x}=\dfrac{2+\sqrt{4\left(1+x^2\right)}}{2x}\le\dfrac{2+\dfrac{4+\left(1+x^2\right)}{2}}{2x}=\dfrac{9+x^2}{4x}\)

Tương tự ta cũng có:

\(\dfrac{1+\sqrt{1+y^2}}{y}\le\dfrac{9+y^2}{4y};\dfrac{1+\sqrt{1+z^2}}{z}\le\dfrac{9+z^2}{4z}\)

Cộng theo vế 3 BĐT trên ta có:

\(\dfrac{1+\sqrt{1+x^2}}{x}+\dfrac{1+\sqrt{1+y^2}}{y}+\dfrac{1+\sqrt{1+z^2}}{z}\le\dfrac{9+x^2}{4x}+\dfrac{9+y^2}{4y}+\dfrac{9+z^2}{4z}\)

\(=\dfrac{9\left(xy+yz+xz\right)+xyz\left(x+y+z\right)}{4xyz}\le\dfrac{9\cdot\dfrac{\left(x+y+z\right)^2}{3}+\left(xyz\right)^2}{4xyz}=xyz\)

Đẳng thức xảy ra khi \(x=y=z=\sqrt{3}\)

Bài 1:

\(\dfrac{2a}{\sqrt{1+a^2}}=\dfrac{2a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)

Sau đó côsi

Tự làm nốt nhé, ra 3/2 đấy. Em học lớp 8 nên cách giải chỉ thế thôi. Câu 2 em chưa làm được

bài này dễ cho xin 1 slot giải bài giờ làm đề cương đã

Lời giải:

\(\frac{1}{2x+y+6}+\frac{1}{2y+z+6}+\frac{1}{2z+x+6}\leq \frac{1}{4}\)

\(\Leftrightarrow \frac{6}{2x+y+6}+\frac{6}{2y+z+6}+\frac{6}{2z+x+6}\leq \frac{3}{2}\)

\(\Leftrightarrow 1-\frac{2x+y}{2x+y+6}+1-\frac{2y+z}{2y+z+6}+1-\frac{2z+x}{2z+x+6}\leq \frac{1}{4}\)

\(\Leftrightarrow A=\frac{2x+y}{2x+y+6}+\frac{2y+z}{2y+z+6}+\frac{2z+x}{2z+x+6}\geq \frac{3}{2}\)

-----------------------

Thật vậy. Áp dụng BĐT Cauchy-Schwarz:

\(A=\frac{(2x+y)^2}{(2x+y)(2x+y+6)}+\frac{(2y+z)^2}{(2y+z)(2y+z+6)}+\frac{(2z+x)^2}{(2z+x)(2z+x+6)}\)

\(\geq \frac{(2x+y+2y+z+2z+x)^2}{ (2x+y)(2x+y+6)+(2y+z)(2y+z+6)+(2z+x)(2z+x+6)}\)

\(\Leftrightarrow A\geq \frac{9(x+y+z)^2}{5(x^2+y^2+z^2)+4(xy+yz+xz)+18(x+y+z)}\)

Ta sẽ cm \( \frac{9(x+y+z)^2}{5(x^2+y^2+z^2)+4(xy+yz+xz)+18(x+y+z)}\geq \frac{3}{2}\)

\(\Leftrightarrow \frac{3(x+y+z)^2}{5(x^2+y^2+z^2)+4(xy+yz+xz)+18(x+y+z)}\geq \frac{1}{2}\)

\(\Leftrightarrow x^2+y^2+z^2+8(xy+yz+xz)\geq 18(x+y+z)\)

\(\Leftrightarrow (x+y+z)^2+6(xy+yz+xz)\geq 18(x+y+z)(*)\)

Theo BĐT AM-GM: \((xy+yz+xz)^2\geq 3xyz(x+y+z)\)

\(\Leftrightarrow (xy+yz+xz)^2\geq 24xyz\Rightarrow xy+yz+xz\geq 2\sqrt{6(x+y+z)}\)

Đặt \(\sqrt{6(x+y+z)}=t\)

Có \((x+y+z)^2+6(xy+yz+xz)\geq \frac{t^4}{36}+12t\geq 18.\frac{t^2}{6}\)

\(\Leftrightarrow \frac{t^3}{36}+12\geq 3t\)

\(\Leftrightarrow t^3-108t+432\geq 0\)

\(\Leftrightarrow (t-6)^2(t+12)\geq 0\) (luôn đúng với mọi \(t\geq 0\) )

Do đó ta có \((*)\), từ \((*)\Rightarrow A\geq \frac{3}{2}\). CM kết thúc

Dấu bằng xảy ra khi \(x=y=z=2\)

Ta có: \(\dfrac{1}{1+x}\ge2-\dfrac{1}{1+y}-\dfrac{1}{1+z}=1-\dfrac{1}{1+y}+1-\dfrac{1}{1+z}\)

\(=\dfrac{y}{1+y}+\dfrac{z}{1+z}=2\sqrt{\dfrac{yz}{\left(1+y\right)\left(1+z\right)}}\)

Tương tự vs 2 bđt còn lại: \(\left\{{}\begin{matrix}\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(1+x\right)\left(1+z\right)}}\\\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\end{matrix}\right.\)

Nhân các vế của 3 bđt trên => ĐPCM

Đề sai nhé, \(\dfrac{z^2}{x+1}\) mới đúng nha

\(\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+3}\left(\text{Svácxơ}\right)\)

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

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)

Ta có: \(x+y+z\ge3\sqrt[3]{xyz}=3\)

\(\Rightarrow x+y+z+3\le2\left(x+y+z\right)\)