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Bài 1:
Ta có:
\(P=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1\)
\(P=\left[\left(a+1\right)\left(a+4\right)\right]\cdot\left[\left(a+2\right)\left(a+3\right)\right]+1\)
\(P=\left(a^2+5a+4\right)\left(a^2+5a+6\right)+1\)
Đặt \(x=a^2+5a+5\) , khi đó:
\(P=\left(a-1\right)\left(a+1\right)+1\)
\(P=a^2-1+1\)
\(P=a^2=\left(x^2-5x+5\right)^2\)
Mà \(a\inℤ\Rightarrow x^2-5x+5\inℤ\)
=> P là số chính phương
\(\left(xy+yz+zx\right)^2+\left(x^2-yz\right)^2+\left(y^2-zx\right)^2+\left(z^2-xy\right)^2=x^2y^2+y^2z^2+z^2x^2+2xyz\left(x+y+z\right)+x^4-2x^2yz+y^2z^2+y^4-2y^2zx+z^2x^2+z^4-2z^2xy+x^2y^2=x^4+y^4+z^4+2\left(x^2y^2+y^2z^2+z^2x^2\right)=\left(x^2+y^2+z^2\right)^2=100^2=10000\)
Từ \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)
\(\Rightarrow\)\(x+y+z=xyz\)
Ta có : \(\sqrt{yz\left(1+x^2\right)}=\sqrt{yz+x^2yz}=\sqrt{yz+x\left(x+y+z\right)}=\sqrt{\left(x+y\right)\left(x+z\right)}\)
Tương tự : \(\sqrt{xy\left(1+z^2\right)}=\sqrt{\left(z+y\right)\left(z+x\right)}\); \(\sqrt{zx\left(1+y^2\right)}=\sqrt{\left(y+z\right)\left(y+x\right)}\)
Nên \(Q=\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\frac{y}{\sqrt{\left(y+z\right)\left(y+x\right)}}+\frac{z}{\sqrt{\left(z+x\right)\left(z+y\right)}}\)
\(Q=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}+\sqrt{\frac{z}{x+z}.\frac{z}{y+z}}\)
Áp dụng BĐT \(\sqrt{A.B}\le\frac{A+B}{2}\left(A,B>0\right)\)
Dấu "=" xảy ra khi A = B :
Ta được :
\(Q\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+x}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{z}{z+y}\right)=\frac{3}{2}\)
Vậy GTLN của \(Q=\frac{3}{2}\)khi \(x=y=z=\sqrt{3}\)
Mình nghĩ bạn viết hơi sai đề bài.
\(x^2+xz-y^2-yz=\left(x^2-y^2\right)+xz-yz=\left(x-y\right)\left(x+y\right)+z\left(x-y\right)=\left(x-y\right)\left(x+y+z\right)\)
Tương tự: \(y^2+xy-z^2-xz=\left(y-z\right)\left(x+y+z\right)\)
\(z^2+yz-x^2-xy=\left(x+y+z\right)\left(z-x\right)\)
Khi đó:
\(P=\frac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}+\frac{1}{\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}+\frac{1}{\left(x-y\right)\left(x+y+z\right)\left(z-x\right)}\)
\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}=0\)
Từ dữ kiện đề bài => x + y + z = xyz
Ta có :
\(\frac{x}{\sqrt{yz\left(1+x^2\right)}}=\frac{x}{\sqrt{yz+xyz.x}}=\frac{x}{\sqrt{yz+x\left(x+y+z\right)}}=\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}\)
\(=\frac{\sqrt{x}}{\sqrt{x+z}}.\frac{\sqrt{x}}{\sqrt{x+y}}\le\frac{1}{2}.\left(\frac{x}{x+z}+\frac{x}{x+y}\right)\)
Tương tự với hai hạng tử còn lại , suy ra
\(Q\le\frac{1}{2}\left(\frac{x}{x+z}+\frac{x}{x+y}\right)+\frac{1}{2}\left(\frac{y}{x+y}+\frac{y}{y+z}\right)+\frac{1}{2}\left(\frac{z}{z+x}+\frac{z}{z+y}\right)=\frac{3}{2}\)
Vậy Max = 3/2 <=> x = y = z
Nguồn : Đinh Đức Hùng
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)