\(VT=\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)=3+\dfrac{x^2+y^2}{z^2}+z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\)

\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}>=2\cdot\sqrt{\dfrac{y^2}{x^2}\cdot\dfrac{x^2}{y^2}}=2\)

=>\(VT>=5+\left(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}\right)+\left(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}\right)+\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)

\(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}>=2\cdot\sqrt{\dfrac{x^2}{z^2}\cdot\dfrac{z^2}{16x^2}}=\dfrac{1}{2}\)

\(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}>=\dfrac{1}{2}\)

và \(\dfrac{1}{x^2}+\dfrac{1}{y^2}>=\dfrac{2}{xy}>=\dfrac{2}{\left(\dfrac{x+y}{2}\right)^2}=\dfrac{8}{\left(x+y\right)^2}\)

=>\(\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)>=\dfrac{15}{16}z^2\cdot\dfrac{8}{\left(x+y\right)^2}=\dfrac{15}{2}\left(\dfrac{z}{x+y}\right)^2=\dfrac{15}{2}\)

=>VT>=5+1/2+1/2+15/2=27/2

Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!

1) a) \(\dfrac{x^2-y^2}{x^3}+y^{^3}.\left(\dfrac{xy-x^2-y^2}{y}.\dfrac{xy}{y-x}\right)\)

\(=\dfrac{x^2-y^2}{x^3}+y^3.\dfrac{x\left(xy-x^2-y^2\right)}{y-x}\)

\(=\dfrac{x^2-y^2}{x^3}+\dfrac{xy^3\left(xy-x^2-y^2\right)}{y-x}\)

\(=\dfrac{-\left(x-y\right)^2\left(x+y\right)+xy^3\left(xy-x^2-y^2\right)}{x^3\left(y-x\right)}\)

Cậu tự thu gọn nốt nhé , tớ sắp đi hok

Bài 2 . Theo giả thiết : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

=> \(\dfrac{yz+xz+xy}{xyz}=\dfrac{1}{x+y+z}\)

=> \(\left(x+y+z\right)\left(yz+zx+xy\right)=xyz\)

=>\(x\left(yz+xz+xy\right)+y\left(yz+xz+xy\right)+z\left(yz+xz+xy\right)-xyz=0\)=> \(\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)

Ta có :

* x = - y

* y = -z

* x = -z

Áp dụng đều này vào phân thức cần CM , ta có :

TH1 . x = -y

\(\dfrac{1}{\left(-y\right)^5}+\dfrac{1}{y^5}+\dfrac{1}{z^5}=\dfrac{1}{\left(-y\right)^5+y^5+z^5}\)

=> \(\dfrac{1}{z^5}=\dfrac{1}{z^5}\), luôn đúng

Tương tự thử với các trường hợp còn lại ta cũng sẽ có được đpcm

Lời giải:

Ta có: \(xy+yz+xz=3xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)

Mà theo BĐT Cauchy-Schwarz: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\)

Do đó: \(3\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3\)

-------

Ta có: \(\text{VT}=x-\frac{xz}{x^2+z}+y-\frac{xy}{y^2+x}+z-\frac{yz}{z^2+y}\)

\(=(x+y+z)-\left(\frac{xy}{y^2+x}+\frac{yz}{z^2+y}+\frac{xz}{x^2+z}\right)\)

\(\geq x+y+z-\frac{1}{2}\left(\frac{xy}{\sqrt{xy^2}}+\frac{yz}{\sqrt{z^2y}}+\frac{xz}{\sqrt{x^2z}}\right)\) (AM-GM)

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

Tiếp tục AM-GM: \(\sqrt{x}+\sqrt{y}+\sqrt{z}\leq \frac{x+1}{2}+\frac{y+1}{2}+\frac{z+1}{2}=\frac{x+y+z+3}{2}\)

Suy ra:

\(\text{VT}\geq x+y+z-\frac{1}{2}.\frac{x+y+z+3}{2}=\frac{3}{4}(x+y+z)-\frac{3}{4}\)

\(\geq \frac{9}{4}-\frac{3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

Ta có đpcm

Dấu bằng xảy ra khi $x=y=z=1$

\(VT=\dfrac{\left(\dfrac{1}{z}\right)^2}{\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\left(\dfrac{1}{x}\right)^2}{\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\left(\dfrac{1}{y}\right)^2}{\dfrac{1}{x}+\dfrac{1}{z}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

Dâu "=" xảy ra khi \(x=y=z\)

\(A=\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)}\)

\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2.\dfrac{x+y+z}{xyz}}\)

Vì x+y+z =0 \(\Rightarrow A=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\) (đpcm)

Đặt \(\left(x;y;z\right)=\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\Rightarrow abc=1\)

\(P=\dfrac{a^2bc}{b+c}+\dfrac{ab^2c}{c+a}+\dfrac{abc^2}{a+b}=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(P=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\)