Ta có:

\(x^2+1=x^2+xy+yz+zx\)

           \(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)

Tương tự:

\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)

\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)

\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

TH1: x,y,z <0

\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)

TH2: x,y,z>0

\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)

Ta có \(1+z^2=xy+yz+zx+z^2\)

\(=y\left(x+z\right)+z\left(x+z\right)\)

\(=\left(x+z\right)\left(y+z\right)\)

CMTT, \(1+x^2=\left(x+y\right)\left(x+z\right)\) và \(1+y^2=\left(x+y\right)\left(y+z\right)\)

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

\(=\sqrt{\left(y+z\right)^2}\) \(=\left|y+z\right|\)

 Tương tự như thế, ta được

\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

 Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.

Lời giải:

$xy+yz+xz=1$
$\Rightarrow x^2+1=x^2+xy+yz+xz=(x+y)(x+z)$

Tương tự: $y^2+1=(y+z)(y+x); z^2+1=(z+x)(z+y)$

Khi đó:

\(\sum \sqrt{\frac{(x^2+1)(y^2+1)}{z^2+1}}=\sum \sqrt{\frac{(x+y)(x+z)(y+x)(y+z)}{(z+x)(z+y)}}=\sum \sqrt{(x+y)^2}\)

$=\sum (x+y)=2(x+y+z)$

Áp dụng bất đẳng thức AM - GM:

\(P\ge3\sqrt[3]{\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\).

Áp dụng bất đẳng thức AM - GM ta có:

\(xy+1=xy+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\ge5\sqrt[5]{\dfrac{xy}{4^4}}\).

Tương tự: \(yz+1\ge5\sqrt[5]{\dfrac{yz}{4^4}};zx+1\ge5\sqrt[5]{\dfrac{zx}{4^4}}\).

Do đó \(\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\ge125\sqrt[5]{\dfrac{\left(xyz\right)^2}{4^{12}}}\)

\(\Rightarrow\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{1}{4^{12}\left(xyz\right)^3}}\).

Mà \(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{8}\)

Nên \(\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{8^3}{4^{12}}}=125\sqrt[5]{\dfrac{1}{2^{15}}}=\dfrac{125}{8}\)

\(\Rightarrow P\ge\dfrac{15}{2}\).

Vậy...

Áp dụng bất đẳng thức AM - GM:

P≥33√(xy+1)(yz+1)(zx+1)xyz.

Áp dụng bất đẳng thức AM - GM ta có:

xy+1=xy+14+14+14+14≥55√xy44.

Tương tự: yz+1≥55√yz44;zx+1≥55√zx44.

Do đó (xy+1)(yz+1)(zx+1)≥1255√(xyz)2412

⇒(xy+1)(yz+1)(zx+1)xyz≥1255√1412(xyz)3.

Mà xyz≤(x+y+z)327=18

Nên  (xy+1)(yz+1)(zx+1)xyz≥1255√83412=1255√1215=1258 

⇒P≥152.

Lời giải:

Từ \(x+y+z=xyz\Rightarrow \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)

Đặt \((\frac{1}{a}, \frac{1}{b}, \frac{1}{c})=(x,y,z)\), trong đó $a,b,c>0$ thì ta có:

\(ab+bc+ac=1\) và cần phải CMR:

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

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Ta có:
\(\frac{\sqrt{(\frac{1}{b^2}+1)(\frac{1}{c^2}+1})-\sqrt{\frac{1}{b^2}+1}-\sqrt{\frac{1}{c^2}+1}}{\frac{1}{bc}}=\sqrt{(b^2+1)(c^2+1)}-b\sqrt{c^2+1}-c\sqrt{b^2+1}\)

\(=\sqrt{(b^2+ab+bc+ac)(c^2+ac+bc+ab)}-b\sqrt{c^2+ac+bc+ab}-c\sqrt{b^2+ab+bc+ac}\)

\(=\sqrt{(b+a)(b+c)(c+a)(c+b)}-b\sqrt{(c+a)(c+b)}-c\sqrt{(b+a)(b+c)}\)

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

Tương tự:

\(\frac{\sqrt{(\frac{1}{c^2}+1)(\frac{1}{a^2}+1})-\sqrt{\frac{1}{c^2}+1}-\sqrt{\frac{1}{a^2}+1}}{\frac{1}{ac}}=(a+c)\sqrt{(b+a)(b+c)}-a\sqrt{(c+a)(c+b)}-c\sqrt{(a+b)(a+c)}(2)\)

\(\frac{\sqrt{(\frac{1}{a^2}+1)(\frac{1}{b^2}+1})-\sqrt{\frac{1}{a^2}+1}-\sqrt{\frac{1}{b^2}+1}}{\frac{1}{ab}}=(a+b)\sqrt{(c+a)(c+b)}-b\sqrt{(a+b)(a+c)}-a\sqrt{(b+c)(b+a)}(3)\)

Từ \((1);(2);(3)\Rightarrow P=(b+c-c-b)\sqrt{(a+b)(a+c)}+(a+c-c-a)\sqrt{(b+a)(b+c)}+(a+b-b-a)\sqrt{(c+a)(c+b)}\)

\(=0\)

Ta có đpcm.

Thế 1=xy+yz+zx vào A ý

1 + y² = xy + yz + xz + y² = (x + y)(y + z)

1 + z² = xy + yz + xz + z² = (x + z)(z + y)

1 + x² = xy + yz + xz + x² = (y + x)(x + z)

Sau khi thay vào và rút gọn ta được

S = x(y + z) + y(x + z) + z(x + y)

S = 2(xy + yz + xz) = 2.1 = 2

Ace Legona

\(P\le\sqrt{3\left(\sum\dfrac{1}{\left(x+y\right)^2+\left(x+1\right)^2+4}\right)}\le\sqrt{3\left(\sum\dfrac{1}{4xy+4x+4}\right)}\)

\(P\le\sqrt{\dfrac{3}{4}\sum\left(\dfrac{1}{xy+x+1}\right)}=\dfrac{\sqrt{3}}{2}\)

\(P_{max}=\dfrac{\sqrt{3}}{2}\) khi \(x=y=z=1\)