\(\left(xy+yz+zx\right)^2\ge3xyz\left(x+y+z\right)=9\Rightarrow xy+yz+zx\ge3\)

\(2\left(x^2+y^2\right)-xy\ge\left(x+y\right)^2-\dfrac{1}{4}\left(x+y\right)^2=\dfrac{3}{4}\left(x+y\right)^2\)

Tương tự và nhân vế với vế:

\(VT\ge\dfrac{27}{64}\left[\left(x+y\right)\left(y+z\right)\left(z+x\right)\right]^2\)

Mặt khác ta có:

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

\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)

\(\ge\left(x+y+z\right)\left(xy+yz+xz\right)-\dfrac{1}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)

\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\ge\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)}.\left(xy+yz+zx\right)\)

\(\Rightarrow VT\ge\dfrac{27}{64}.\dfrac{64}{81}.3\left(xy+yz+zx\right)^3\ge3^3=27\) (đpcm)

Đặt \(\hept{\begin{cases}2x+y+z=4a\\2y+x+z=4b\\2z+x+y=4c\end{cases}\Rightarrow}\hept{\begin{cases}x=3a-b-c\\y=3b-c-a\\z=3c-a-b\end{cases}}\)thay vào biểu thức đó

\(\Rightarrow\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}\)

\(=\frac{3a-b-c}{4a}+\frac{3b-c-a}{4b}+\frac{3c-a-b}{4c}\)

\(=\frac{3}{4}-\frac{b-c}{4a}+\frac{3}{4}-\frac{c-a}{4b}+\frac{3}{4}-\frac{a-b}{4c}\)

\(=\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\)

Áp dụng BĐT sau: \(\frac{a}{b}+\frac{b}{a}\ge2\Rightarrow\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\ge6\)

\(\Leftrightarrow\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\ge\frac{6}{4}\)

\(\Leftrightarrow\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\le\frac{3}{4}\)

Từ đó ta có: \(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}\le\frac{3}{4}\)(đpcm).

Dấu "=" xảy ra <=> x=y=z.

Theo Cauchy Schwarz:

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

Tương tự:

\(\frac{y}{2y+z+x}\le\frac{1}{4}\left(\frac{y}{y+x}+\frac{y}{y+z}\right);\frac{z}{2z+y+x}\le\frac{1}{4}\left(\frac{z}{z+y}+\frac{z}{z+x}\right)\)

Cộng lại:

\(D\le\frac{3}{4}\left(đpcm\right)\)

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

\(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2xz}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{xz+2yz}\)

\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\) 

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)

Ta có: \(\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}=\frac{x^4}{xy+2xz}+\frac{y^4}{yz+2yx}+\frac{z^4}{zx+2zy}\)

Áp dụng BĐT Cauchy Schwarz, ta có:

\(=\frac{x^4}{xy+2xz}+\frac{y^4}{yz+2yx}+\frac{z^4}{zx+2zy}\ge\frac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\frac{1}{3}\)

=> ĐPCM

Dấu "=" xảy ra khi: \(x=y=z=\frac{1}{\sqrt{3}}\)

Áp dụng BĐT Cosi cho 2 số dương, ta có:

\(\frac{9x^3}{y+2z}+x\left(y+2z\right)\ge6x^2;\frac{9y^3}{z+2x}+y\left(z+2x\right)\ge6y^2;\frac{9z^3}{x+2y}+z\left(x+2y\right)\ge6z^3\)

Lại có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\Rightarrow x^2+y^2+z^2\ge xy+yz+zx\)

Do đó \(\frac{9x^3}{y+2z}+\frac{9y^3}{z+2x}+\frac{9z^3}{x+2y}+3\left(xy+yz+zx\right)\ge6\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow\frac{9x^3}{y+2z}+\frac{9y^3}{z+2x}+\frac{9z^3}{x+2y}\ge6\left(x^2+y^2+z^2\right)-3\left(xy+yz+zx\right)\ge3\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}\ge\frac{x^2+y^2+z^2}{3}=\frac{1}{3}\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{1}{\sqrt{3}}\)

Bạn CM x=y=z=1

Sau đó bạn thế số vào và bạn sẽ tính đc phân số là 3/6 rút gọn là 1/2

Cuối cùng bạn sẽ kết luận:

Vì 1/2 ≤ 1/2

Nên ...(biểu thức)...≤1/2

CM x=y=z kiểu gì vậy???

** Bạn lưu ý lần sau viết đề bằng công thức toán!

Đề cần sửa thành $\leq \frac{4}{3}$

Lời giải:

Áp dụng BĐT AM-GM và Cauchy-Schwarz:

\(\frac{1}{2x^2+y^2+z^2}=\frac{1}{(x^2+z^2)+(x^2+y^2)}\leq \frac{1}{2xy+2xz}=\frac{1}{2}.\frac{1}{xy+xz}\leq \frac{1}{8}\left(\frac{1}{xy}+\frac{1}{xz}\right)\)

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế suy ra:

\(\sum \frac{1}{2x^2+y^2+z^2}\leq \frac{1}{4}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)=\frac{x+y+z}{4xyz}\) $(1)$

Mặt khác:

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\Rightarrow 4xyz=xy+yz+xz$

$\Rightarrow 16x^2y^2z^2=(xy+yz+xz)^2\geq 3xyz(x+y+z)$ (theo BĐT AM-GM)

$\Rightarrow x+y+z\leq \frac{16}{3}xyz (2)$

Từ $(1);(2)\Rightarrow \sum \frac{1}{2x^2+y^2+z^2}\leq \frac{4}{3}$ 

Dấu "=" xảy ra khi $x=y=z=\frac{3}{4}$

\(\dfrac{1}{2x^2+y^2+z^2}=\dfrac{1}{x^2+y^2+x^2+z^2}\le\dfrac{1}{2xy+2xz}\le\dfrac{1}{8}\left(\dfrac{1}{xy}+\dfrac{1}{xz}\right)\)

Tương tự: \(\dfrac{1}{x^2+2y^2+z^2}\le\dfrac{1}{8}\left(\dfrac{1}{xy}+\dfrac{1}{yz}\right)\) ; \(\dfrac{1}{x^2+y^2+2z^2}\le\dfrac{1}{8}\left(\dfrac{1}{xz}+\dfrac{1}{yz}\right)\)

Cộng vế:

\(VT\le\dfrac{1}{4}\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)\le\dfrac{1}{4}.\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2=\dfrac{4}{3}\)

Đề bài sai