Xem lại cái đề đi Tuyển. Hình như giá trị nhỏ nhất của cái biểu thức dưới còn lớn hơn là 1 thì làm sao bài đó có giá trị x, y, z thỏa được mà bảo tính A.

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.

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.

Gọi cái thiệt gớm đó là P

Ta có:

\(xy+yz+zx=xyz\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)

Ta có:

\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64y}+\dfrac{1+y}{64x}\ge3\sqrt[3]{\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}.\dfrac{1+x}{64y}.\dfrac{1+y}{64x}}=\dfrac{3}{16z}\)

\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{64x}-\dfrac{1}{64y}-\dfrac{1}{32}\left(1\right)\)

Tương tự ta cũng có:

\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{64y}-\dfrac{1}{64z}-\dfrac{1}{32}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{64z}-\dfrac{1}{64x}-\dfrac{1}{32}\left(3\right)\end{matrix}\right.\)

Từ (1), (2), (3) ta được

\(P\ge\dfrac{3}{16}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)

\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)

Dấu = xảy ra khi \(x=y=z=3\)

Đặt cái ban đầu là P

Ta có: \(xy+yz+zx=xyz\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)

Ta lại có:

\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64x}+\dfrac{1+y}{64y}\ge\dfrac{3}{16z}\)

\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{32}-\dfrac{1}{64x}-\dfrac{1}{64y}\left(1\right)\)

Tương tự ta có:

\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{32}-\dfrac{1}{64y}-\dfrac{1}{64z}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{32}-\dfrac{1}{64z}-\dfrac{1}{64x}\left(3\right)\end{matrix}\right.\)

Từ (1), (2), (3) ta có:

\(P\ge\dfrac{3}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)

\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)

Dấu = xảy ra khi \(x=y=z=3\)

Chứng minh gì bạn?

Lời giải:

Ta thấy: \(xy+yz+xz=1\)

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

Do đó:

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

Hoàn toàn tt:

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

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

Cộng theo vế:

\(S=x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2\)

Lời giải:

Ta thấy: xy+yz+xz=1

⇒⎧⎪⎨⎪⎩1+y2=xy+yz+xz+y2=(y+z)(y+x)1+x2=xy+yz+xz+x2=(x+y)(x+z)1+z2=xy+yz+xz+z2=(z+x)(z+y)

Do đó:

x√(y2+1)(z2+1)1+x2=x√(y+x)(y+z)(z+x)(z+y)(x+y)(x+z)=x√(y+z)2=x(y+z)

Hoàn toàn tt:

y√(x2+1)(z2+1)y2+1=y(x+z) 

z√(x2+1)(y2+1)z2+1=z(x+y)

Cộng theo vế:

S=x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2