pt 1) x=y=z  Cosi 3 số 

a) xy(x + y) + yz(y + z) + xz(z + x) + 3xyz

= xy(X + y + z)  + yz(x + y + z) + xz(X + y + z)

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

b) xy(x + y) - yz(y + z) - xz(z - x)

= x²y + xy² - y²z - yz² - xz² + x²z

= x²(y + z) - yz(y + z) + x(y² - z²)

= x²(y + z) - yz(y + z) + x(y + z)(y - z)

= (y + z)(x² - yz + xy - xz)

= (y + z)[x(x + y) - z(x + y)]

= (y + z)(x + y)(x - z)

c) x(y² - z²) + y(z² - x²) + z(x² - y²)

 = x(y - z)(y + z) + yz² - yx² + x²z - y²z

= x(y - z)(y + z) - yz(y - z) - x²(y - z)

= (y - z)((xy + xz - yz - x²)

= (y - z)[x(y - x) - z(y - x)]

= (y - z)(x - z)(y -x) 

Lời giải:

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

Đặt \((a,b,c)=\left(\frac{1}{x}; \frac{1}{y}; \frac{1}{z}\right)\Rightarrow a+b+c=1\)

BĐT cần chứng minh trở thành:

\(P=\frac{c^3}{(a+1)(b+1)}+\frac{a^3}{(b+1)(c+1)}+\frac{b^3}{(c+1)(a+1)}\geq \frac{1}{16}(*)\)

Thật vậy, áp dụng BĐT Cauchy ta có:

\(\frac{c^3}{(a+1)(b+1)}+\frac{a+1}{64}+\frac{b+1}{64}\geq 3\sqrt[3]{\frac{c^3}{64^2}}=\frac{3c}{16}\)

\(\frac{a^3}{(b+1)(c+1)}+\frac{b+1}{64}+\frac{c+1}{64}\geq 3\sqrt[3]{\frac{a^3}{64^2}}=\frac{3a}{16}\)

\(\frac{b^3}{(c+1)(a+1)}+\frac{c+1}{64}+\frac{a+1}{64}\geq 3\sqrt[3]{\frac{b^3}{64^2}}=\frac{3b}{16}\)

Cộng theo vế các BĐT trên và rút gọn :

\(\Rightarrow P+\frac{a+b+c+3}{32}\geq \frac{3(a+b+c)}{16}\)

\(\Leftrightarrow P+\frac{4}{32}\geq \frac{3}{16}\Leftrightarrow P\geq \frac{1}{16}\)

Vậy \((*)\) được chứng minh. Bài toán hoàn tất.

Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\Leftrightarrow x=y=z=3\)

Đa thức trên tương đương với đa thức:

\(\left(xy\left(x+y\right)+xyz\right)+\left(yz\left(y+z\right)+xyz\right)+\left(xz\left(x+z\right)+xyz\right)\)

=\(xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+y+z\right)\)

=\(\left(x+y+z\right)\left(xy+yz+xz\right)\)

xy(x + y) + yz( y + z )+ zx( z + x ) + 3xyz

=xy(x + y) + xyz + yz(y + z) + xyz + xz(x + z)+xyz

=zy(x + y + z) + yz(x + y + z) + xz(x + y + z)

=(x + y + z)(xy + yz + zx)

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