+\(10=x+3y=x+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}\ge10\sqrt[10]{\frac{1}{3^9}x.y^9}\)

\(=\frac{10}{3}.\sqrt[10]{3}.\sqrt[10]{xy^9}\)

\(\Rightarrow xy^9\le3^9\)

+\(\frac{1}{\sqrt{x}}+\frac{27}{\sqrt{3y}}=\frac{1}{\sqrt{x}}+\frac{3}{\sqrt{3y}}+\frac{3}{\sqrt{3y}}+.....+\frac{3}{\sqrt{3y}}\)

\(\ge10\sqrt[10]{\frac{3^9}{\sqrt{3^9x.y^9}}}\ge10\sqrt[10]{\frac{3^9}{\sqrt{3^9.3^9}}}=10\)

Dấu "=" xảy ra khi và chỉ khi \(x=1;y=3\)

x + 25 = 64

x         = 64 - 25

x         = 39

Vậy x = 39

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

dùng buniacosky với x+3y<10 là dc

Giải ra

\(\text{Cho 3 số dương x, y, z thỏa mãn }x+y+z=3\)

\(\text{Chứng minh rằng }T=\dfrac{x}{x+\sqrt{3x+yz}}+\dfrac{y}{y+\sqrt{3y+xz}}+\dfrac{z}{z+\sqrt{3z+xy}}\le1\)

➤➤➤Chứng minh:

➢ Áp dụng bất đẳng thức AM - GM

\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}\left(\text{vì }x+y+z=3\right)=\dfrac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}\le\dfrac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}}=\dfrac{x}{x+\sqrt{xz}+\sqrt{xy}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

➢ Tương tự:

\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

\(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

➢ Công vế theo vế 3 bất đẳng thức cùng chiều

\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}+\dfrac{y}{y+\sqrt{3y+xz}}+\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)

➢ \(\text{Đẳng thức xảy ra khi }x=y=z=1\)

➤ \(Max_T=1\Leftrightarrow x=y=z=1\)

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

\(\Rightarrow\sqrt{x}+3\sqrt{3y}\le10\)

Đặt \(P=\frac{1}{\sqrt{x}}+\frac{27}{\sqrt{3y}}\)

\(P=\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{27}{\sqrt{3y}}+3\sqrt{3y}-\left(\sqrt{x}+3\sqrt{3y}\right)\)

\(P\ge2+18-10=10\)

"="<=>x=1;y=3

\(x+\sqrt{3x+yz}=x+\sqrt{x\left(x+y+z\right)+yz}=x+\sqrt{\left(x+y\right)\left(z+x\right)}\ge x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}\)

\(=x+\sqrt{xz}+\sqrt{xy}=\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Tương tự:

\(\dfrac{y}{y+\sqrt{3y+zx}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng vế với vế ta có đpcm