Ta có: \(2x+3y=7\Leftrightarrow\dfrac{x}{3}+\dfrac{y}{2}=\dfrac{7}{6}\)

\(3x^2+5y^2=\dfrac{\left(\dfrac{x}{3}\right)^2}{\dfrac{1}{27}}+\dfrac{\left(\dfrac{y}{2}\right)^2}{\dfrac{1}{20}}\ge\dfrac{\left(\dfrac{x}{3}+\dfrac{y}{2}\right)^2}{\dfrac{1}{27}+\dfrac{1}{20}}=\dfrac{\left(\dfrac{7}{6}\right)^2}{\dfrac{1}{27}+\dfrac{1}{20}}=\dfrac{735}{47}\) (đpcm)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\dfrac{70}{47}\\y=\dfrac{63}{47}\end{matrix}\right.\)

 đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)

\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)

BBDT AM-GM 

\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)

theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)

vì \(x^2+y^2\ge2xy\)

\(y^2+z^2\ge2yz\)

\(x^2+z^2\ge2xz\)

\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)

\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)

\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)

dấu"=" xảy ra<=>x=y=z=1/3

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

\(VT\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+5xy+5yz+5zx}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+3\left(xy+yz+zx\right)}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(x+y+z\right)^2}=\dfrac{1}{2}\)

4 không chia hết cho 49. Bạn xem lại đề xem lỗi ở đâu.

Ta có: \(\dfrac{1}{\sqrt{x}}+\dfrac{27}{\sqrt{3y}}=\dfrac{1}{\sqrt{x}}+\dfrac{81}{3\sqrt{3y}}\ge\dfrac{\left(1+9\right)^2}{\sqrt{x}+3\sqrt{3y}}=\dfrac{100}{\sqrt{x}+3\sqrt{3y}}\) (1)

Áp dụng BĐT của Cô-si ta có:

    \(\sqrt{x}=\sqrt{1.x}\le\dfrac{1+x}{2};3\sqrt{3y}\le\dfrac{9+3y}{2}\)

\(\Rightarrow\left(1\right)\ge\dfrac{100}{\dfrac{1+x}{2}+\dfrac{9+3y}{2}}=\dfrac{100}{\dfrac{10+x+3y}{2}}\ge\dfrac{100}{\dfrac{10+10}{2}}=\dfrac{100}{10}=10\)

Dấu "=" xảy ra ⇔ x=1;y=3

\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2xz+2yz=z^2+\left(x+y\right)^2+2z\left(x+y\right)=36\)

áp dụng BĐT cosi : 

\(z^2+\left(x+y\right)^2\ge2z\left(x+y\right)\)

<=> \(z^2+\left(x+y\right)^2+2z\left(x+y\right)\ge4z\left(x+y\right)=36< =>z\left(x+y\right)\ge9\)

ta lại có \(\dfrac{x+y}{xyz}=\dfrac{x}{xyz}+\dfrac{y}{xyz}=\dfrac{1}{yz}+\dfrac{1}{xz}\) áp dụng BĐT buhihacopxki dạng phân thức => \(\dfrac{1}{yz}+\dfrac{1}{xz}\ge\dfrac{4}{yz+xz}=\dfrac{4}{z\left(x+y\right)}\ge\dfrac{4}{9}\left(đpcm\right)\)

dấu bằng xảy ra khi \(\left[{}\begin{matrix}yz=xz< =>x=y\\x+y+z=6\\z^2=\left(x+y\right)^2\end{matrix}\right.< =>\left[{}\begin{matrix}x+y+z=6\\z=2x=2y\end{matrix}\right.< =>\left[{}\begin{matrix}x=y=\dfrac{3}{2}\\z=3\end{matrix}\right.\)

\(\dfrac{x+y}{xyz}=\dfrac{1}{yz}+\dfrac{1}{xz}\).

Áp dụng bất đẳng thức Cauchy-Schawrz dạng Engel:

\(\dfrac{1}{yz}+\dfrac{1}{xz}\ge\dfrac{4}{z\left(x+y\right)}\)     (1).

Áp dụng bất đẳng thức Cauchy cho hai số dương z và x+y, ta có:

\(z\left(x+y\right)\le\left(\dfrac{x+y+z}{2}\right)^2=9\). Suy ra, \(\dfrac{4}{z\left(x+y\right)}\ge\dfrac{4}{9}\)     (2).

Từ (1) và (2), suy ra \(\dfrac{x+y}{xyz}\ge\dfrac{4}{9}\) (đpcm).

Dấu "=" xảy ra khi và chỉ khi \(\dfrac{1}{yz}=\dfrac{1}{xz}\) và \(z=x+y\).