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Ta sẽ chứng minh
\(\sqrt{x^2+1}+2\sqrt{x}\le\frac{2+\sqrt{2}}{2}\left(x+1\right)\)
\(\Leftrightarrow\left(\sqrt{x^2+1}+2\sqrt{x}\right)^2\le\frac{3+2\sqrt{2}}{2}\left(x+1\right)^2\)
\(\Leftrightarrow\frac{1+2\sqrt{2}}{2}\left(x^2+1\right)-4\sqrt{x\left(x^2+1\right)}+\left(2\sqrt{2}-1\right)x\ge0\)
\(\Leftrightarrow\left(\sqrt{x^2+1}-\sqrt{2x}\right)\left(\frac{1+2\sqrt{2}}{2}\sqrt{x^2+1}-\frac{4-\sqrt{2}}{2}\sqrt{x}\right)\ge0\)
BĐT trên luôn đúng do \(x^2+1\ge2x\)
Vậy ta có:\(\text{∑}\sqrt{x^2+1}+2\sqrt{x}\le\text{∑}\frac{2+\sqrt{2}}{2}\left(x+1\right)\le6+3\sqrt{2}\)
Đẳng thức xảy ra khi x=y=z=1
\(\sqrt{x-y+z}=\sqrt{x}-\sqrt{y}+\sqrt{z}\)
Điều kiện tự làm nhé
\(\Leftrightarrow x-y+z=x+y+z+2\left(\sqrt{xz}-\sqrt{xy}-\sqrt{yz}\right)\)
\(\Leftrightarrow y+\sqrt{xz}-\sqrt{xy}-\sqrt{yz}\)
\(\Leftrightarrow\left(\sqrt{z}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=y\\z=y\end{cases}}\)
\(M^2=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2xy}{\sqrt{yz}}+\frac{2yz}{\sqrt{zx}}+\frac{2xz}{\sqrt{yz}}=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2x\sqrt{y}}{\sqrt{z}}+\frac{2y\sqrt{z}}{\sqrt{x}}+\frac{2z\sqrt{x}}{\sqrt{y}}\)
Áp dụng bđt Cô-si: \(\frac{x^2}{y}+\frac{x\sqrt{y}}{\sqrt{z}}+\frac{x\sqrt{y}}{\sqrt{z}}+z\ge4\sqrt[4]{\frac{x^2}{y}.\frac{x\sqrt{y}}{\sqrt{z}}.\frac{x\sqrt{y}}{\sqrt{z}}.z}=4x\)
tương tự \(\frac{y^2}{z}+\frac{y\sqrt{z}}{\sqrt{x}}+\frac{y\sqrt{z}}{\sqrt{x}}+x\ge4y\);\(\frac{z^2}{x}+\frac{z\sqrt{x}}{\sqrt{y}}+\frac{z\sqrt{x}}{\sqrt{y}}+y\ge4z\)
=>\(M^2+x+y+z\ge4\left(x+y+z\right)\Rightarrow M^2\ge3\left(x+y+z\right)\ge3.12=36\Rightarrow M\ge6\)
Dấu "=" xảy ra khi x=y=z=4
Vậy minM=6 khi x=y=z=4
2. Xem tại đây
1. \(P=\frac{1}{\sqrt{x.1}}+\frac{1}{\sqrt{y.1}}+\frac{1}{\sqrt{z.1}}\)
\(\ge\frac{1}{\frac{x+1}{2}}+\frac{1}{\frac{y+1}{2}}+\frac{1}{\frac{z+1}{2}}\)
\(=\frac{2}{x+1}+\frac{2}{y+1}+\frac{2}{z+1}\ge\frac{2.\left(1+1+1\right)^2}{x+y+z+3}=\frac{18}{3+3}=3\)
Đẳng thức xảy ra \(\Leftrightarrow x=y=z=1\)
1 ) có cách theo cosi đó
áp dụng cosi cho 3 số dương ta có \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}+x\ge3\sqrt[3]{\frac{1}{\sqrt{x}}\times\frac{1}{\sqrt{x}}\times x}=3\sqrt[3]{1}=3\)(1)
\(\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{y}}+y\ge3\)(2)
\(\frac{1}{\sqrt{z}}+\frac{1}{\sqrt{z}}+z\ge3\)(3)
cộng các vế của (1),(2),(3), đc \(2\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)+\left(x+y+z\right)\ge9\Rightarrow2P+3\ge9\Rightarrow P\ge3\)
minP=3 khi x=y=z=1
Đặt \(\sqrt{x^2+y^2}=c;\sqrt{y^2+z^2}=a;\sqrt{z^2+x^2}=b\)
Ta có:
\(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
\(\ge\frac{x^2}{\sqrt{2\left(y^2+z^2\right)}}+\frac{y^2}{\sqrt{2\left(z^2+x^2\right)}}+\frac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)
\(=\frac{1}{2\sqrt{2}}\left(\frac{c^2+b^2-a^2}{a}+\frac{a^2+c^2-b^2}{b}+\frac{b^2+a^2-c^2}{c}\right)\)
\(\ge\frac{1}{2\sqrt{2}}\left(\frac{\left(2a+2b+2c\right)^2}{2\left(a+b+c\right)}-2018\right)=\frac{1009}{\sqrt{2}}\)
1.\(N=x^2+\frac{1000}{x}+\frac{1000}{x}\ge3\sqrt[3]{\frac{x^2.1000.1000}{x^2}}\)
\(\Rightarrow N\ge300\)
Dấu "=" xảy ra \(\Leftrightarrow x^3=1000\Leftrightarrow x=10\)
2.\(P=\left(5x+\frac{12}{x}\right)+\left(3y+\frac{16}{y}\right)\ge2\sqrt{60}+2\sqrt{48}=4\sqrt{15}+8\sqrt{3}\)
Dấu "=" xảy ra \(\Leftrightarrow5x=\frac{12}{x};3y=\frac{16}{y}\Leftrightarrow x=\sqrt{\frac{12}{5}};y=\frac{4\sqrt{3}}{3}\)
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