giúp mình với cho x+y+z=3 Tìm GTLN xy/(x+3y+2z) + yz/(y+3z+2x) + zx/(z+3x+2y)

*) tìm giá trị lớn nhất: từ giả thiết \(\hept{\begin{cases}0\le x\le1\\0\le y\le1\end{cases}\Leftrightarrow\hept{\begin{cases}x^3\le x^2\\y^3\le y^2\end{cases}\Leftrightarrow}x^3+y^3\le x^2+y^2=1}\)

maxA=1 \(\Leftrightarrow\hept{\begin{cases}x^3=x^2\\y^3=y^2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0;y=1\\x=1;y=0\end{cases}}}\)

*) tìm giá trị nhỏ nhất \(\left(x+y\right)^2\le2\left(x^2+y^2\right)=1\Rightarrow x+y\le\sqrt{2}\Rightarrow\frac{x+y}{\sqrt{2}}\le1\)

do đó \(x^3+y^3\ge\frac{\left(x^3+y^3\right)\left(x+y\right)}{\sqrt{2}}\)theo bđt Bunhiacopxki

\(\left(x^3+y^3\right)\left(x+y\right)=\left[\left(\sqrt{x^3}\right)^2+\left(\sqrt{y^3}\right)^2\right]\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2\right]\)

\(\ge\left(\sqrt{x^3}\cdot\sqrt{x}+\sqrt{y^3}\cdot\sqrt{y}\right)^2=x^2+y^2=1\)

vậy minA=\(\frac{1}{\sqrt{2}}\Leftrightarrow x=y=\frac{\sqrt{2}}{2}\)

\(A=x^2+x\sqrt{3}+1=x^2+2x.\frac{\sqrt{3}}{2}+\frac{3}{4}+\frac{1}{4}\)

                                          \(=\left(x+\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)

\(\text{Dấu "=" xảy ra }\Leftrightarrow x+\frac{\sqrt{3}}{2}=0\Leftrightarrow x=-\frac{\sqrt{3}}{2}\)

\(\text{Vậy}\)\(x=-\frac{\sqrt{3}}{2}\)\(\text{thì}\)\(A_{min}=\frac{1}{4}\)

\(A=x^2+2.\frac{1}{2}\sqrt{3}x+\frac{3}{4}+\frac{1}{4}\)

\(=\left(x+\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\)

vậy GTNN của A là 1/4. Dấu bằng xảy ra khi \(x=\frac{-\sqrt{3}}{2}\)

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

\(\frac{x+2}{x\sqrt{x}+1}+\frac{\sqrt{x}-1}{x-\sqrt{x}+1}-\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}< 1\)

\(\frac{x+2}{\sqrt{x^3}+1}+\frac{\sqrt{x}-1}{x-\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}< 1\)

\(\frac{x+2+x-1-x+\sqrt{x}-1}{\sqrt{x^3}+1}< 1\)

\(\frac{x+\sqrt{x}}{\sqrt{x^3}+1}< 1\Leftrightarrow\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(\Leftrightarrow\frac{\sqrt{x}}{x-\sqrt{x}+1}< 1\)

Vậy tự kết luận

Auto giải thích thêm cái chỗ \(\frac{\sqrt{x}}{x-\sqrt{x}+1}< 1\Leftrightarrow\sqrt{x}< x-\sqrt{x}+1\)

Ta có: \(x\sqrt{x}+1>0\)

\(\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)>0\)

Vì \(x>0\) và  \(x\ne1\)

Do đó: \(\sqrt{x}+1>0\) và  \(x-\sqrt{x}+1>0\)

\(\sqrt{x}< x-\sqrt{x}+1\)( chuyển vế qua )

\(\left(\sqrt{x^2}-1\right)^2>0\)hay  \(x-2\sqrt{x}+1>0\)

Nên \(\sqrt{x}< x-\sqrt{x}+1\) ( luôn đúng )

\(\sqrt{3-\sqrt{3+x}}=x\)

3-\(\sqrt{3+x}\)=x²

3+x-\(\sqrt{3+x}\)+\(\frac{1}{4}\)=x²+x+\(\frac{1}{4}\)

(\(\sqrt{3+x}\)-\(\frac{1}{4}\))²=(x+\(\frac{1}{2}\))²

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{3+x}-\frac{1}{2}=x+\frac{1}{2}\\\sqrt{3+x}-\frac{1}{2}=-x-\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{3+x}=x+1\\\sqrt{3+x}=-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3+x=x^2+2x+1\\3+x=x^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+x-2=0\\x^2-x+\frac{1}{4}=\frac{13}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)\left(x+2\right)=0\\\left(x-\frac{1}{2}\right)^2=\frac{13}{4}\end{cases}}\)

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