a) ĐK \(x\ge1\)

với \(x\ge1\Rightarrow\hept{\begin{cases}\sqrt{x-1}\ge0\\\sqrt{5+x}\ge\sqrt{6}\end{cases}\Rightarrow\sqrt{x-1}+\sqrt{5+x}\ge\sqrt{6}}\)

dâu = xảy ra <=>x=1

b)Dặt ...=A

Ta có A=\(\frac{2}{9}x+\frac{1}{2x}+\frac{2}{9}y+\frac{1}{2y}+\frac{7}{9}\left(x+y\right)\)

Áp dụng BĐT cô-si, ta có \(\frac{2}{9}x+\frac{1}{2x}\ge\frac{2}{3}\)

tương tự có \(\frac{2}{9}y+\frac{1}{2y}\ge\frac{2}{3}\)

Mà \(x+y\ge3\Rightarrow\frac{7}{9}\left(x+y\right)\ge\frac{7}{3}\)

=>\(A\ge\frac{2}{3}+\frac{2}{3}+\frac{7}{3}=\frac{11}{3}\)

Dấu = xảy ra <=>\(x=y=\frac{3}{2}\)

^_^

b) Nó ko > = 11/3 =))

Sau vài phút cố gắng thì khẳng định đề bài của em bị sai

Đề này còn có lý, lần sau chú ý đọc kĩ đề trước khi đăng lên, tránh làm mất thời gian vô ích:

\(\left|x-2y\right|\le\dfrac{1}{\sqrt{x}}\Rightarrow1\ge\sqrt{x}\left|x-2y\right|\Rightarrow1\ge x\left(x-2y\right)^2\)

\(\Rightarrow1\ge x^3-4x^2y+4xy^2\)

Tương tự: \(\dfrac{1}{\sqrt{y}}\ge\left|y-2x\right|\Rightarrow1\ge y^3-4xy^2+4xy^2\)

Cộng vế:

\(\Rightarrow2\ge x^3+y^3=\dfrac{1}{2}\left(x^3+x^3+1\right)+\left(y^3+1+1\right)-\dfrac{5}{2}\ge\dfrac{1}{2}.3x^2+3y-\dfrac{3}{2}=\dfrac{3}{2}\left(x^2+2y\right)-\dfrac{5}{2}\)

\(\Rightarrow\dfrac{3}{2}\left(x^2+2y\right)\le\dfrac{9}{2}\Rightarrow x^2+2y\le3\)

Q=\(\left(1+\dfrac{a}{x}\right)\left(1+\dfrac{a}{y}\right)\left(1+\dfrac{a}{z}\right)\)

\(Q=\left(\dfrac{x+a}{x}\right)\left(\dfrac{y+a}{y}\right)\left(\dfrac{z+a}{z}\right)\)\

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

=\(\dfrac{\left(2x+y+z\right)\left(2y+x+z\right)\left(2z+x+y\right)}{xyz}\)

ÁP dụng BĐT cô si

\(2x+y+z=x+x+y+z\ge4\sqrt[4]{x^2yz}\)

\(2y+x+z=y+y+x+z\ge4\sqrt[4]{y^2xy}\)

\(2z+y+x=z+z+x+y\ge4\sqrt[4]{z^2xy}\)

=> Q\(\ge\dfrac{64.\sqrt[4]{x^4y^4z^4}}{xyz}=64\)

=> MinQ=64 khi x=y=z=a/3

pn oi nhieu the nay ai ma giai cho het dc

bài lớp mấy mà nhìn ghê quá zật bạn..................Nhìu quá

sao mà trả ời hết đc

ko biert lam kho qua

Ta có \(\frac{1}{P}=\frac{\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)^2}{x^3y^3}=\frac{x+yz}{y}\cdot\frac{y+zx}{x}\cdot\frac{\left(z+xy\right)^2}{x^2y^2}\)

\(=\left(\frac{x}{y}+z\right)\left(\frac{y}{x}+z\right)\left(\frac{z}{xy}+1\right)^2=\left[1+\left(\frac{x}{y}+\frac{x}{y}\right)z+x^2\right]\left(\frac{z}{xy}+1\right)^2\ge\left(1+2x+x^2\right)\)\(\left[\frac{4x}{\left(x+y\right)^2}+1\right]^2\)\(=\left(z+1\right)^2\left[\frac{4z}{\left(z-1\right)^2}+1\right]^2=\left[\frac{4z\left(z+1\right)}{\left(z-1\right)^2}+1\right]^2=\left[6+\frac{12}{z-1}+\frac{8}{\left(z-1\right)^2}+z-1\right]^2\)

\(=\left[6+\frac{12}{z-1}+\frac{3\left(z-1\right)}{4}+\frac{8}{\left(z-1\right)^2}+\frac{z-1}{8}+\frac{z-1}{8}\right]\)

Áp dụng BĐT Cosi ta có:

\(\frac{1}{P}\ge\left[6+2\sqrt{\frac{12}{z-1}\cdot\frac{3\left(z-1\right)}{3}}+3\sqrt[3]{\frac{8}{\left(z-1\right)^2}\cdot\frac{z-1}{8}\cdot\frac{z-1}{8}}\right]^2=\frac{729}{4}\)

\(\Rightarrow P\le\frac{4}{729}\). dấu "=" xảy ra <=> \(\hept{\begin{cases}x=y=2\\z=5\end{cases}}\)

khong lay so 1 nho nha

\(\sqrt{x+2+2\sqrt{x+1}}+\sqrt{x+2-2\sqrt{ }x+1}=\frac{x+5}{2}\)\(\frac{x+5}{2}\)

Xét \(x^2+\sqrt{1+x^2}\)ta có:

\(x^2\ge0\)

nên \(1+x^2\ge1\)

\(\Rightarrow\sqrt{1+x^2}\ge\sqrt{1}=1\)

\(\Rightarrow x^2+\sqrt{1+x^2}\ge1\)

Tương tự ta có: 

\(y^2+\sqrt{1+y^2}\ge1\)

Do đó: \(\left(x^2+\sqrt{1+x^2}\right)\left(y^2+\sqrt{1+y^2}\right)\ge1\)

Dấu bằng xảy ra khi \(x=0;y=0\)

Khi đó \(x+y=0\left(ĐPCM\right)\)