\(x+y+z=xyz\Leftrightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)=2^2-2.1=2\) (đpcm)

\(\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\)

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

\(=x.\left(\dfrac{x+y+z}{y+z}\right)+y.\left(\dfrac{x+y+z}{x+z}\right)+z.\left(\dfrac{x+y+z}{x+y}\right)-\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)-\left(x+y+z\right)=\left(x+y+z\right)-\left(x+y+z\right)=0\)

\(\sum\dfrac{1}{x}\cdot\sum\dfrac{x}{y^2}\ge\sum^2\dfrac{1}{x}\)(bunhia)

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

Dâu "=" xảy ra khi \(x=y=z\)

đúng rùi đó

Sai

Lời giải:

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

Bài toán tương đương với: Cho \(a,b,c>0\) và \(abc=1\). CMR

\(a^2+b^2+c^2\geq a+b+c\)

Thật vậy.

Áp dụng BĐT AM-GM: \(a+b+c\geq 3\sqrt[3]{abc}=3\sqrt[3]{1}=3(1)\)

Theo hệ quả của BĐT Am-Gm:

\(a^2+b^2+c^2\geq ab+bc+ac\Rightarrow 3(a^2+b^2+c^2)\geq a^2+b^2+c^2+2(ab+bc+ac)\)

\(\Rightarrow a^2+b^2+c^2\geq \frac{(a+b+c)^2}{3}\)

Kết hợp với \((1)\Rightarrow a^2+b^2+c^2\geq a+b+c\)

Do đó ta có đpcm

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

Phân số cuối cùng chắc em ghi nhầm

\(\dfrac{x}{y+z+t}+\dfrac{y+z+t}{9x}\ge2\sqrt{\dfrac{x\left(y+z+t\right)}{9x\left(y+z+t\right)}}=\dfrac{2}{3}\)

Tương tự:

\(\dfrac{y}{z+t+x}+\dfrac{z+t+x}{9y}\ge\dfrac{2}{3}\)

\(\dfrac{z}{t+x+y}+\dfrac{t+x+y}{9z}\ge\dfrac{2}{3}\)

\(\dfrac{t}{x+y+z}+\dfrac{x+y+z}{9t}\ge\dfrac{2}{3}\)

Đồng thời:

\(\dfrac{8}{9}\left(\dfrac{y+z+t}{x}+\dfrac{z+t+x}{y}+\dfrac{t+x+y}{z}+\dfrac{x+y+z}{t}\right)\)

\(\ge\dfrac{8}{9}\left(\dfrac{3\sqrt[3]{yzt}}{x}+\dfrac{3\sqrt[3]{ztx}}{y}+\dfrac{3\sqrt[3]{txy}}{z}+\dfrac{3\sqrt[3]{xyz}}{t}\right)\)

\(\ge\dfrac{8}{3}.4\sqrt[4]{\dfrac{\sqrt[3]{yzt}.\sqrt[3]{ztx}.\sqrt[3]{txy}.\sqrt[3]{xyz}}{xyzt}}=\dfrac{32}{3}\)

Cộng vế:

\(VT\ge4.\dfrac{2}{3}+\dfrac{32}{3}=\dfrac{40}{3}\)

Dấu "=" xảy ra khi \(x=y=z=t\)

Áp dụng bđt phụ \(\dfrac{1}{A+B}\le\dfrac{1}{4}\left(\dfrac{1}{A}+\dfrac{1}{B}\right)\forall A,B>0\)

\(\dfrac{1}{2x+y+z}=\dfrac{1}{\left(x+y\right)+\left(x+z\right)}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\) Tương tự: \(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\Rightarrow\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=1\)

Dấu bằng xảy ra \(\Leftrightarrow x=y=z=\dfrac{3}{4}\)

\(\dfrac{1}{x+x+y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

Tương tự: \(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)\) ; \(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\)

Cộng vế với vế:

\(VT\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{3}{4}\)