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+\(10=x+3y=x+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}\ge10\sqrt[10]{\frac{1}{3^9}x.y^9}\)
\(=\frac{10}{3}.\sqrt[10]{3}.\sqrt[10]{xy^9}\)
\(\Rightarrow xy^9\le3^9\)
+\(\frac{1}{\sqrt{x}}+\frac{27}{\sqrt{3y}}=\frac{1}{\sqrt{x}}+\frac{3}{\sqrt{3y}}+\frac{3}{\sqrt{3y}}+.....+\frac{3}{\sqrt{3y}}\)
\(\ge10\sqrt[10]{\frac{3^9}{\sqrt{3^9x.y^9}}}\ge10\sqrt[10]{\frac{3^9}{\sqrt{3^9.3^9}}}=10\)
Dấu "=" xảy ra khi và chỉ khi \(x=1;y=3\)
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
Lời giải:
Áp dụng BĐT SVac-xơ:
\(\frac{1}{\sqrt{x}}+\frac{27}{\sqrt{3y}}=\frac{1}{\sqrt{x}}+\frac{9}{\sqrt{3y}}+\frac{9}{\sqrt{3y}}+\frac{9}{\sqrt{3y}}\geq \frac{(1+3+3+3)^2}{\sqrt{x}+3\sqrt{3y}}\)
\(\Leftrightarrow \frac{1}{\sqrt{x}}+\frac{27}{\sqrt{3y}}\geq \frac{100}{x+3\sqrt{3y}}(1)\)
Áp dụng BĐT Bunhiacopxky:
\((x+3y)(1+9)\geq (\sqrt{x}+3\sqrt{3y})^2\)
\(\Rightarrow \sqrt{x}+3\sqrt{3y}\leq \sqrt{10(x+3y)}\leq 10(2)\) do \(x+3y\leq 10\)
Từ \((1);(2)\Rightarrow \frac{1}{\sqrt{x}}+\frac{27}{\sqrt{3y}}\geq \frac{100}{x+3\sqrt{3y}}\geq \frac{100}{10}=10\) (đpcm)
Dấu bằng xảy ra khi \(\frac{\sqrt{x}}{1}=\frac{\sqrt{3y}}{3}; x+3y=10\Rightarrow x=1;y=3\)
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
Cách khác:
\(\frac{\left(x+y\right)^2}{2}+\frac{\left(x+y\right)}{4}\ge2xy+\frac{x+y}{4}\)
\(=\frac{4xy+x+4xy+y}{4}=\frac{x\left(4y+1\right)+y\left(4x+1\right)}{4}\)
\(\ge\frac{4x\sqrt{y}+4y\sqrt{x}}{4}=x\sqrt{y}+y\sqrt{x}\)
Dấu = xảy ra khi \(x=y=\frac{1}{4}\)
\(\frac{1}{2}\left(x+y\right)\left(x+y+\frac{1}{2}\right)=\frac{1}{2}\left(x+y\right)\left(x+\frac{1}{4}+y+\frac{1}{4}\right)\)
Áp dụng bất đẳng thức cauchy:
\(x+y\ge2\sqrt{xy}\)
\(x+\frac{1}{4}\ge2\sqrt{\frac{x}{4}}=\sqrt{x}\)
\(y+\frac{1}{4}\ge2\sqrt{\frac{y}{4}}=\sqrt{y}\)
do đó \(VT\ge\frac{1}{2}.2.\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)=x\sqrt{y}+y\sqrt{x}\)(đpcm)
Dấu = xảy ra khi \(x=y=\frac{1}{4}\)
ta có 3x + yz = x2 + xy + yz + zx = (x+y)(x+z)
do đó:
\(\frac{x}{x+\sqrt{3x+yz}}=\frac{x\left(\sqrt{x^2+xy+yz+zx}-x\right)}{\left(\sqrt{x^2+xy+yz+zx}+x\right)\left(\sqrt{x^2+xy+yz+zx}-x\right)}\)
= \(\frac{x\left(\sqrt{\left(x+y\right)\left(x+z\right)}-x\right)}{xy+yz+zx}\le\frac{x\left(\frac{x+y+x+z}{2}-x\right)}{xy+yz+zx}\)\(\le\frac{x\left(y+z\right)}{2\left(xy+yz+zx\right)}\)
tương tự với 2 số hạng còn lại nên ta được: P\(\le\)1. đpcm