Gt\(\Leftrightarrow\left(x+\sqrt{x^2+2}\right)\left(x-\sqrt{x^2+2}\right)\left(y-1+\sqrt{y^2-2y+3}\right)=2\left(x-\sqrt{x^2+2}\right)\)

\(\Leftrightarrow-2\left(y-1+\sqrt{y^2-2y+3}\right)=2\left(x-\sqrt{x^2+2}\right)\)

\(\Leftrightarrow x-\sqrt{x^2+2}+y-1+\sqrt{y^2-2y+3}=0\) (*)

\(\left(x+\sqrt{x^2+2}\right)\left(y-1+\sqrt{y^2-2y+3}\right)=2\)

\(\Leftrightarrow\left(x+\sqrt{x^2+2}\right)\left(y-1+\sqrt{y^2-2y+3}\right)\left(y-1-\sqrt{y^2-2y+3}\right)=2\left(y-1-\sqrt{y^2-2y+3}\right)\)

\(\Leftrightarrow\left(x+\sqrt{x^2+2}\right).-2=2\left(y-1-\sqrt{y^2+2y+3}\right)\)

\(\Leftrightarrow y-1-\sqrt{y^2+2y+3}+x+\sqrt{x^2+2}=0\) (2*)

Cộng vế với vế của (*) và (2*) => \(2x+2y-2=0\)

\(\Leftrightarrow x+y=1\)

\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\)

\(\Leftrightarrow x^3+y^3+3xy=1\)

Ta có:`(x+sqrt{x^2+2})(sqrt{x^2+2}-x)=2`

`<=>sqrt{x^2+2}-x=y-1+sqrt{y^2-2y+3}`

`<=>sqrt{x^2+2}-sqrt{y^2-2y+3}=x+y-1(1)`

CMTT:`sqrt{y^2-2y+3}-(y-1)=x+sqrt{x^2+2}`

`<=>sqrt{y^2-2y+3}-y+1=x+sqrt{x^2+2}`

`<=>sqrt{y^2-2y+3}-sqrt{x^2+2}=x+y-1(2)`

Cộng từng vế (1)(2) ta có:

`2(x+y-1)=0`

`<=>x+y-1=0`

`<=>x+y=1`

`<=>(x+y)^3=1`

`<=>x^3+y^3+3xy(x+y)=1`

`<=>x^3+y^3+3xy=1`(do `x+y=1`)

ĐKXĐ: x,y >1

\(\sqrt{x^2+5}+\sqrt{x-1}+x^2=\sqrt{y^2+5}+\sqrt{y-1}+y^2\\ \)

\(\Leftrightarrow\sqrt{x^2+5}-\sqrt{y^2+5}+\left(\sqrt{x-1}-\sqrt{y-1}\right)+x^2-y^2=0\)

\(\Leftrightarrow\frac{\left(\sqrt{x^2+5}-\sqrt{y^2+5}\right).\left(\sqrt{x^2+5}+\sqrt{y^2+5}\right)}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\frac{\left(\sqrt{x-1}-\sqrt{y-1}\right).\left(\sqrt{x-1}+\sqrt{y-1}\right)}{\sqrt{x-1}+\sqrt{y-1}}+\left(x^2-y^2\right)=0\)

\(\Leftrightarrow\frac{\left(x^2+5\right)-\left(y^2+5\right)}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\frac{\left(x-1\right)-\left(y-1\right)}{\sqrt{x-1}+\sqrt{y-1}}+\left(x^2-y^2\right)=0\)

\(\Leftrightarrow\frac{x^2-y^2}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\frac{x-y}{\sqrt{x-1}+\sqrt{y-1}}+\left(x^2-y^2\right)=0\)

\(\Leftrightarrow\left(x-y\right).\left(\frac{x+y}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\frac{1}{\sqrt{x-1}+\sqrt{y-1}}+x+y\right)=0\)

\(\Rightarrow x-y=0\Leftrightarrow x=y\)

Giả sử x=y

Khi đó:

\(\sqrt{x^2+5}+\sqrt{x-1}+x^2\)

\(=\sqrt{y^2+5}+\sqrt{x-1}+y^2\)

Luôn đúng 

Vậy ta suy ra đpcm

\(GT\Rightarrow\left(\sqrt{2+x^2}-x\right)\left(\sqrt{2+x^2}+x\right)\left(\sqrt{2+y^2}+y\right)=2\left(\sqrt{2+x^2}+x\right)\)

\(\Leftrightarrow2\left(\sqrt{2+y^2}+y\right)=2\left(\sqrt{2+x^2}+x\right)\)

\(\Leftrightarrow\sqrt{2+x^2}+x-\sqrt{2+y^2}-y=0\)

\(\Leftrightarrow\dfrac{\left(x-y\right)\left(x+y\right)}{\sqrt{2+x^2}+\sqrt{2+y^2}}+\left(x-y\right)=0\)

TH1:\(x-y=0\Leftrightarrow x=y\left(đpcm\right)\)

TH2: \(x+y+\sqrt{2+x^2}+\sqrt{2+y^2}=0\)

Ta có: \(x\ge-\sqrt{x^2}\); \(y\ge-\sqrt{y^2}\)

\(\Rightarrow x+y+\sqrt{2+x^2}+\sqrt{2+y^2}\ge\sqrt{2+x^2}-\sqrt{x^2}+\sqrt{2+y^2}-\sqrt{y^2}>0\)

Do vậy TH2 không có x,y tm

Vậy ta có đpcm

\(\left(x;y;z\right)=\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\Rightarrow ab+bc+ca=2020\)

BĐT trở thành:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+a+b+c+\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le\dfrac{2020.2021}{abc}\)

\(\Leftrightarrow\dfrac{ab+bc+ca}{abc}+a+b+c+\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le\dfrac{2020.2021}{abc}\)

\(\Leftrightarrow a+b+c+\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le\dfrac{2020^2}{abc}\)

Ta có: \(\sqrt{2020+a^2}=\sqrt{ab+bc+ca+a^2}=\sqrt{\left(a+b\right)\left(a+c\right)}\le\dfrac{1}{2}\left(2a+b+c\right)\)

Tương tự:...

\(\Rightarrow\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le2\left(a+b+c\right)\)

\(\Rightarrow a+b+c+\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le3\left(a+b+c\right)\)

Nên ta chỉ cần chứng minh:

\(3\left(a+b+c\right)\le\dfrac{2020^2}{abc}=\dfrac{\left(ab+bc+ca\right)^2}{abc}\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\) (hiển nhiên đúng)

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

biết làm rồi

VẬy bạn giải ra cho mọi người xem được ko?

Lớn hơn hoặc bằng kí hiệu trong Latex là \geq nha!

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)