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Áp dụng BĐT Bu-nhi-a, ta có \(\left(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\right)^2\le3\left(2x+2y+2z\right)=6\)
=> A\(\le\sqrt{6}\)
dấu = xảy ra <=> x=y=z=1/3
\(1=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{z}\right)+\frac{1}{2}\left(\frac{y}{z}+\frac{z}{x}\right)+\frac{1}{2}\left(\frac{z}{x}+\frac{x}{y}\right)\)
\(\ge\sqrt{\frac{x}{y}.\frac{y}{z}}+\sqrt{\frac{y}{z}.\frac{z}{x}}+\sqrt{\frac{z}{x}.\frac{x}{y}}=VP\) (rút gọn lại thôi:v)
Ta có \(\sqrt{1+x^2}+\sqrt{2x}\le\sqrt{2}\left(x+1\right)\)
\(\sqrt{1+y^2}+\sqrt{2y}\le\sqrt{2}\left(y+1\right)\)
\(\sqrt{1+z^2}+\sqrt{2z}\le\sqrt{2}\left(z+1\right)\)
\(\Rightarrow\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2}+\sqrt{2x}+\sqrt{2y}+\sqrt{2z}\le\sqrt{2}\left(x+y+z+3\right)\le6\sqrt{2}\)
Ta lại có \(\sqrt{x}+\sqrt{y}+\sqrt{z}\le\sqrt{3\left(x+y+z\right)}\le3\)
Theo đề bài ta có
\(\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2}+3\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)
\(\le6\sqrt{2}+\left(3-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\le3\sqrt{2}+9\)
Dấu = xảy ra khi x = y = z = 1
Áp dụng bđt côsi ta có:
\(\hept{\begin{cases}\sqrt{\left(x+y\right)4}\le\frac{x+y+4}{2}\left(1\right)\\\sqrt{\left(z+y\right)4}\le\frac{y+z+4}{2}\left(2\right)\\\sqrt{\left(z+x\right)4}\le\frac{z+x+4}{2}\left(3\right)\end{cases}}\)
Lấy \(\left(1\right)+\left(2\right)+\left(3\right)\)ta được:
\(2P\le x+y+z+6=12\)
\(\Leftrightarrow p\le6\)
Dấu"="xảy ra \(\Leftrightarrow x=y=z=2\)
Vậy \(P_{max}=6\)\(\Leftrightarrow x=y=z=2\)
Ta có: \(\left(\sqrt{x+y}\right)^2=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)
\(\Leftrightarrow\)\(x+y=x+y-2z+2\sqrt{\left(x-z\right)\left(y-z\right)}\)
\(\Leftrightarrow2z=2\sqrt{\left(x-z\right)\left(y-z\right)}\)
Theo giả thiết, ta có:
theo giả thiết, ta có: \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0\Rightarrow\frac{1}{z}-\frac{1}{x}=\frac{1}{y}\)\(\Rightarrow\frac{x-z}{zx}=\frac{1}{y}\Rightarrow x-z=\frac{zx}{y}\)
Tương tự, ta có: \(y-z=\frac{zy}{x}\)
Do đó: \(2\sqrt{\left(x-z\right)\left(y-z\right)}=2\sqrt{\frac{zx}{y}.\frac{zy}{x}}=2z\) (1)
ta có: \(\left(\sqrt{x+y}\right)^2=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)
\(\Leftrightarrow2z=2\sqrt{\left(x-z\right)\left(y-z\right)}\)(2)
Thay (2) vào (1) ta thấy (2) luôn đúng
Suy ra ĐPCM
Bài 1: Áp dụng BĐT AM-GM ta có:
\(1+x\ge2\sqrt{x}\)
\(x+y\ge2\sqrt{xy}\)
\(y+1\ge2\sqrt{y}\)
Cộng theo vế 3 BĐT trên ta có:
\(2\left(1+x+y\right)\ge2\left(\sqrt{x}+\sqrt{xy}+\sqrt{y}\right)\)
\(1+x+y\ge\sqrt{x}+\sqrt{xy}+\sqrt{y}\Leftrightarrow VT\ge VP\)
Đẳng thức xảy ra khi \(\hept{\begin{cases}1+x=2\sqrt{x}\\x+y=2\sqrt{xy}\\y+1=2\sqrt{y}\end{cases}}\Rightarrow x=y=1\)
Khi đó \(S=x^{2013}+y^{2013}=1^{2013}+1^{2013}=2\)
Bài 2: Vì \(\hept{\begin{cases}x,y,z\in\left[-1;3\right]\\x+y+z=3\end{cases}}\) nên
\(0\le\left(x+1\right)\left(y+1\right)\left(z+1\right)+\left(3-x\right)\left(3-y\right)\left(3-z\right)\)
\(\Leftrightarrow0\le4\left(xy+yz+xz\right)-8\left(x+y+z\right)+28\)
\(\Leftrightarrow0\le2\left(xy+yz+xz\right)+2\)
\(\Leftrightarrow x^2+y^2+z^2\le x^2+y^2+z^2+2\left(xy+yz+xz\right)+2\)
\(\Leftrightarrow x^2+y^2+z^2\le\left(x+y+z\right)^2+2\)
\(\Leftrightarrow x^2+y^2+z^2\le3^2+2=9+2=11\)
Ta có : \(A=\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\)
\(\Rightarrow A^2=\left(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\right)^2\)
Theo BĐT Bu - nhi - a - cốp - xki ta có :
\(A^2=\left(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\right)^2\le\left(1^2+1^2+1^2\right)\left[2\left(x+y+z\right)\right]=3.2=6\)
\(\Rightarrow A=\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\le\sqrt{6}\) khi \(x=y=z=\dfrac{1}{3}\)
Thank you very much!!!!!!, my friend.