Question

[văn dũng]

AE GIÚP VỚI. CẦN CÁC AE CAO THỦ GIÚP. DEADLINE: 9H SÁNG 19/12/2020

Answer 1

[Hồng Phúc]

5.

\(A=\dfrac{x}{x+\sqrt{x+yz}}+\dfrac{y}{y+\sqrt{y+zx}}+\dfrac{z}{z+\sqrt{z+xy}}\)

\(=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}+\dfrac{y}{y+\sqrt{y\left(x+y+z\right)+zx}}+\dfrac{z}{z+\sqrt{z\left(x+y+z\right)+xy}}\)

\(=\dfrac{x}{x+\sqrt{x^2+xy+yz+zx}}+\dfrac{y}{y+\sqrt{y^2+xy+yz+zx}}+\dfrac{z}{z+\sqrt{z^2+xy+yz+zx}}\)

\(=\dfrac{x\left(\sqrt{x^2+xy+yz+zx}-x\right)}{xy+yz+zx}+\dfrac{y\left(\sqrt{y^2+xy+yz+zx}-y\right)}{xy+yz+zx}+\dfrac{z\left(\sqrt{z^2+xy+yz+zx}-z\right)}{xy+yz+zx}\)

\(=\dfrac{x\sqrt{\left(x+y\right)\left(z+x\right)}-x^2}{xy+yz+zx}+\dfrac{y\sqrt{\left(x+y\right)\left(y+z\right)}-y^2}{xy+yz+zx}+\dfrac{z\sqrt{\left(z+x\right)\left(y+z\right)}-z^2}{xy+yz+zx}\)

Áp dụng BĐT \(ab\le\dfrac{a^2+b^2}{2}\) và BĐT \(a^2+b^2+c^2\ge ab+bc+ca\)

\(A=\dfrac{x\sqrt{\left(x+y\right)\left(z+x\right)}-x^2}{xy+yz+zx}+\dfrac{y\sqrt{\left(x+y\right)\left(y+z\right)}-y^2}{xy+yz+zx}+\dfrac{z\sqrt{\left(z+x\right)\left(y+z\right)}-z^2}{xy+yz+zx}\)

\(=\dfrac{x\sqrt{\left(x+y\right)\left(z+x\right)}+y\sqrt{\left(x+y\right)\left(y+z\right)}+z\sqrt{\left(z+x\right)\left(y+z\right)}-\left(x^2+y^2+z^2\right)}{xy+yz+zx}\)

\(\le\dfrac{x.\dfrac{2x+y+z}{2}+y.\dfrac{x+2y+z}{2}+z.\dfrac{x+y+2z}{2}-\left(x^2+y^2+z^2\right)}{xy+yz+zx}\)

\(=\dfrac{xy+yz+zx}{xy+yz+zx}=1\)

\(maxA=1\Leftrightarrow x=y=z=\dfrac{1}{3}\)

Answer 2

[Hồng Phúc]

1.

a, \(A=(\dfrac{1}{2};2];B=[\dfrac{2}{3};+\infty)\)

b, \(A\cap B=\left[\dfrac{2}{3};2\right];A\cup B=\left(\dfrac{1}{2};+\infty\right)\)