\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=4\)

\(\Leftrightarrow\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=1\)

\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)

Tương tự: \(b+1=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\)

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

\(VT=\sum\dfrac{\sqrt{a}}{a+1}=\sum\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)

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

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

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

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

\(\Rightarrow VT=VP\) (đpcm)

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

\(P=\sqrt{\dfrac{yz}{xy+xz}}+\sqrt{\dfrac{zx}{xy+yz}}+\sqrt{\dfrac{xy}{yz+zx}}\)

\(P=\dfrac{2yz}{2\sqrt{yz\left(xy+xz\right)}}+\dfrac{2zx}{2\sqrt{zx\left(xy+yz\right)}}+\dfrac{2xy}{2\sqrt{xy\left(yz+zx\right)}}\)

\(P\ge\dfrac{2yz}{xy+yz+zx}+\dfrac{2zx}{xy+yz+zx}+\dfrac{2xy}{xy+yz+zx}=2\)

Dấu "=" không xảy ra nên \(P>2\)

`sqrta+sqrtb+sqrtc=2`

`<=>(sqrta+sqrtb+sqrtc)^2=4`

`<=>a+b+c+2sqrt{ab}+2sqrt{bc}+2sqrt{ca}=4`

`<=>2sqrt{ab}+2sqrt{bc}+2sqrt{ca}=4-(a+b+c)=4-2-2`

`<=>sqrt{ab}+sqrt{bc}+sqrt{ca}=1`

`=>a+1=a+sqrt{ab}+sqrt{bc}+sqrt{ca}=sqrta(sqrta+sqrtb)+sqrtc(sqrta+sqrtb)=(sqrta+sqrtb)(sqrta+sqrtc)`

Tương tự:`b+1=(sqrtb+sqrta)(sqrtb+sqrtc)`

`c+1=(sqrtc+sqrta)(sqrtc+sqrtb)`

`=>VT=sqrta/((sqrta+sqrtb)(sqrta+sqrtc))+sqrtb/((sqrtb+sqrta)(sqrtb+sqrtc))+sqrtc/((sqrtc+sqrta)(sqrtc+sqrtb))`

`=>VT=(sqrta(sqrtb+sqrtc)+sqrtb(sqrtc+sqrta)+sqrtc(sqrta+sqrtb))/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=(sqrt{ab}+sqrt{ac}+sqrt{bc}+sqrt{ab}+sqrt{ac}+sqrt{bc})/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=(2(sqrt{ab}+sqrt{bc}+sqrt{ca}))/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=2/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=2/\sqrt{[(sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta)]^2}`

`=2/\sqrt{(sqrta+sqrtb)(sqrta+sqrtc)(sqrtb+sqrta)(sqrtb+sqrtc)(sqrtc+sqrta)(sqrtc+sqrtb)}`

`=2/\sqrt{(1+a)(1+b)(1+c)}=>đpcm`

Lời giải:
\(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\)

\(\Rightarrow (\sqrt{a}+\sqrt{b}+\sqrt{c})^2=4\)

\(\Leftrightarrow a+b+c+2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=4\)

\(\Leftrightarrow \sqrt{ab}+\sqrt{bc}+\sqrt{ac}=\frac{4-(a+b+c)}{2}=1\)

\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})\)

Tương tự:

$b+1=(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})$
$c+1=(\sqrt{c}+\sqrt{a})(\sqrt{c}+\sqrt{b})$

Khi đó:

\(A=\left[\frac{\sqrt{a}}{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})}+\frac{\sqrt{b}}{(\sqrt{b}+\sqrt{a})(\sqrt{b}+\sqrt{c})}+\frac{\sqrt{c}}{(\sqrt{c}+\sqrt{a})(\sqrt{c}+\sqrt{b})}\right]\sqrt{(a+1)(b+1)(c+1)}\)

\(\frac{\sqrt{a}(\sqrt{b}+\sqrt{c})+\sqrt{b}(\sqrt{c}+\sqrt{a})+\sqrt{c}(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})}.\sqrt{(\sqrt{a}+\sqrt{b})^2(\sqrt{b}+\sqrt{c})^2(\sqrt{c}+\sqrt{a})^2}\)

\(=\frac{2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})}.(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})\)

\(=2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=2\)

Ta chứng minh 2 bất đẳng thức phụ sau: với x, y, z dương thì:

\(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\left(1\right)\)

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)\ge\left(1+\sqrt[3]{xyz}\right)^3\left(2\right)\)

+ Chứng minh BĐT (1), sử dụng BĐT AM - GM:

\(x^4+x^4+y^4+z^4\ge4x^2yz\)

\(y^4+y^4+x^4+z^4\ge4xy^2z\)

\(z^4+z^4+x^4+y^4\ge4xyz^2\)

Cộng dồn lại ta có: \(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\)

+ Chứng minh BĐT (2). Ta có:

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)=1+x+y+z+xy+yz+xyz\ge1+3\sqrt[3]{xyz}+3\sqrt[3]{x^2y^2z^2}+xyz=\left(1+\sqrt[3]{xyz}\right)^3\)

Bây giờ ta quay lại chứng minh BĐT ở đề.

BĐT cần chứng minh tương đương với BĐT sau:

\(\sqrt[4]{\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4}\ge\sqrt[4]{3}+\dfrac{\sqrt[4]{243}}{2+abc}\)

\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Sử dụng BĐT (1) ta có:

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

Sử dụng BĐT (2) và BĐT AM - GM ta có:

\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^3\left(3+\dfrac{3}{\sqrt[3]{abc}}\right)\)

\(\Rightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(1+\dfrac{1}{\sqrt[3]{abc.1.1}}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Vậy BĐT đã được chứng minh. Đẳng thức xảy ra <=> a = b = c.

từ giả thiết ,ta có:\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=4\)\(\Leftrightarrow a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)=4\)

\(\Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=1\)---> thay 1= vào ...

Đề bài hình như bị sai em, thay điểm rơi ko thỏa mãn

Biểu thức là \(a+b+\sqrt{2\left(a+c\right)}\) mới đúng

em cũng nghĩ thế mới dùng đc BDT AM-GM 3 số đúng ko thầy :)