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Áp dụng bất đẳng thức Cauchy ta có :
\(VT=\frac{1}{\sqrt{a}}+\frac{3}{\sqrt{b}}+\frac{8}{\sqrt{3c+2a}}\)
\(=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2}{\sqrt{b}}+\frac{8}{\sqrt{3c+2a}}\)
\(\ge\frac{4}{\sqrt{a}+\sqrt{b}}+\frac{2\left(1+2\right)^2}{\sqrt{3c+2a}+\sqrt{b}}\)
\(=\frac{4}{\sqrt{a}+\sqrt{b}}+\frac{\left(1+2\right)^2}{\sqrt{3c+2a}+\sqrt{b}}+\frac{\left(1+2\right)^2}{\sqrt{3c+2a}+\sqrt{b}}\)
\(\ge\frac{\left(1+2+1+2+2\right)^2}{2\sqrt{3c+2a}+3\sqrt{b}+\sqrt{a}}\)
\(\ge\frac{64}{\sqrt{\left(1+2^2+3\right)\left(a+2a+3c+3b\right)}}\)
\(=\frac{64}{\sqrt{24\left(a+c+b\right)}}=\frac{16\sqrt{2}}{\sqrt{3\left(a+b+c\right)}}=VP\)
\(\sqrt{\frac{ab+2c^2}{1+ab-c^2}}=\sqrt{\frac{ab+2c^2}{a^2+b^2+ab}}=\frac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(a^2+b^2+ab\right)}}\ge\frac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\ge\frac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)
Tương tự: \(\sqrt{\frac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\) ; \(\sqrt{\frac{ca+2b^2}{1+ac-b^2}}\ge ca+2b^2\)
Cộng vế với vế:
\(VT\ge2\left(a^2+b^2+c^2\right)+ab+bc+ca=2+ab+bc+ca\)
1)
\(2a+\frac{4}{a}+\frac{16}{a+2}=\left(a+\frac{4}{a}\right)+\left[\left(a+2\right)+\frac{16}{a+2}\right]-2\ge4+8-2=10\)
Dấu "=" xảy ra khi a=2
2)
\(\hept{\begin{cases}\sqrt{a\left(1-4a\right)}=\frac{1}{2}\sqrt{4a\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4a+1-4a}{2}=\frac{1}{4}\\\sqrt{b\left(1-4b\right)}=\frac{1}{2}\sqrt{4\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4b+1-4b}{2}=\frac{1}{4}\\\sqrt{c\left(1-4c\right)}=\frac{1}{2}\sqrt{4c\left(1-4c\right)}\le\frac{1}{2}\cdot\frac{4c+1-4c}{2}=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\sqrt{a\left(1-4a\right)}+\sqrt{b\left(1-4b\right)}+\sqrt{c\left(1-4c\right)}\le\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{8}\)
\(15\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+30\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=40\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2007\)
\(\Leftrightarrow15\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=40\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2007\)
\(\Leftrightarrow15\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le\frac{40}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+2007\)
\(\Leftrightarrow\frac{5}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le2007\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{\frac{6021}{5}}\)
Ta có:
\(5a^2+2ab+2b^2=4a^2+2ab+b^2+a^2+b^2\ge4a^2+2ab+b^2+2ab=\left(2a+b\right)^2\)
\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}=\frac{1}{a+a+b}+\frac{1}{b+b+c}+\frac{1}{c+c+a}\)
\(\Rightarrow P\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow P\le\frac{1}{3}\sqrt{\frac{6021}{5}}\)
Dấu "=" xảy ra khi \(a=b=c=3\sqrt{\frac{5}{6021}}\)
Mẫu thức như vầy thì tìm max còn được chứ tìm min sao nổi bạn?
2.
Áp dụng bất đẳng thức Bunhiacopxki :
\(\left(1+9^2\right)\left(x^2+\frac{1}{x^2}\right)\ge\left(x+\frac{9}{x}\right)^2\)
\(\Leftrightarrow82\cdot\left(x^2+\frac{1}{x^2}\right)\ge\left(x+\frac{9}{x}\right)^2\)
\(\Leftrightarrow\sqrt{82}\cdot\sqrt{x^2+\frac{1}{x^2}}\ge x+\frac{9}{x}\)
Tương tự ta cũng có :
\(\sqrt{82}\cdot\sqrt{y^2+\frac{1}{y^2}}\ge y+\frac{9}{y}\)
\(\sqrt{82}\cdot\sqrt{z^2+\frac{1}{z^2}}\ge z+\frac{9}{z}\)
Cộng theo vế của các bất đẳng thức ta được :
\(\sqrt{82}\cdot\left(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\right)\ge x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\)
\(\Leftrightarrow\sqrt{82}\cdot P\ge x+\frac{9}{x}+y+\frac{9}{y}+z+\frac{9}{z}\)(1)
Mặt khác áp dụng bất đẳng thức Cauchy ta có :
\(x+\frac{9}{x}+y+\frac{9}{y}+z+\frac{9}{z}=81x+\frac{9}{x}+81y+\frac{9}{y}+81z+\frac{9}{z}-80x-80y-80z\)
\(\ge2\sqrt{\frac{81x\cdot9}{x}}+2\sqrt{\frac{81y\cdot9}{y}}+2\sqrt{\frac{81z\cdot9}{z}}-80\left(x+y+z\right)\)
\(\ge2\sqrt{729}+2\sqrt{729}+2\sqrt{729}-80\cdot1\)
\(=82\) (2)
Từ (1) và (2) suy ra \(\sqrt{82}\cdot P\ge82\)
\(\Leftrightarrow P\ge\sqrt{82}\) ( đpcm )
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)
1.
Áp dụng bất đẳng thức Cauchy :
\(\frac{a^2+1}{a}+\frac{b^2+1}{b}+\frac{c^2+1}{c}\)
\(=a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\)
\(=9a+\frac{1}{a}+9b+\frac{1}{b}+9c+\frac{1}{c}-8a-8b-8c\)
\(\ge2\sqrt{\frac{9a}{a}}+2\sqrt{\frac{9b}{b}}+2\sqrt{\frac{9c}{c}}-8\left(a+b+c\right)\)
\(\ge3\cdot2\sqrt{9}-8=10\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
P=\(\frac{a-1}{\sqrt{b-1}}\sqrt{\frac{b-2\sqrt{b}+1}{a^2-2a+1}}=\frac{a-1}{\sqrt{b-1}}\sqrt{\frac{\left(\sqrt{b}-1\right)^2}{\left(a-1\right)^2}}=\frac{a-1}{\sqrt{b-1}}.(\frac{\sqrt{b}-1}{a-1})=\frac{\sqrt{b}-1}{\sqrt{b-1}}\)