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\(a^2=\dfrac{\sqrt{2}}{4}\left(1-a\right)\)
\(\Rightarrow a^4=\dfrac{1}{8}\left(1-a\right)^2\)
\(\Rightarrow a^4+a+1=\dfrac{1}{8}\left(1-a\right)^2+a+1=\dfrac{1}{8}\left(a^2+6a+9\right)=\dfrac{1}{8}\left(a+3\right)^2\)
\(\Rightarrow\sqrt{a^4+a+1}-a^2=\sqrt{\dfrac{1}{8}\left(3+a\right)^2}-a^2=\dfrac{\sqrt{2}}{4}\left(a+3\right)-\dfrac{\sqrt{2}}{4}\left(1-a\right)=\dfrac{\sqrt{2}}{2}\left(a+1\right)\)
\(\Rightarrow\dfrac{a+1}{\sqrt{a^4+a+1}-a^2}=\dfrac{a+1}{\dfrac{\sqrt{2}}{2}\left(a+1\right)}=\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Khi x=16 thì \(A=\dfrac{4+1}{4-1}=\dfrac{5}{3}\)
b: \(P=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{x-4}=\dfrac{x+\sqrt{x}-2}{x-4}=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)
c: \(P=A\cdot B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=1+\dfrac{3}{\sqrt{x}-2}\)
Để P lớn nhất thì căn x-2=1
=>căn x=3
=>x=9
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(P=\sqrt{\dfrac{yz}{x^2+1}}+\sqrt{\dfrac{zx}{y^2+1}}+\sqrt{\dfrac{xy}{z^2+1}}\)
\(P=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}+\sqrt{\dfrac{zx}{y^2+xy+yz+zx}}+\sqrt{\dfrac{xy}{z^2+xy+yz+zx}}\)
\(P=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\dfrac{zx}{\left(y+z\right)\left(x+y\right)}}+\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\)
\(P\le\dfrac{1}{2}\left(\dfrac{y}{x+y}+\dfrac{z}{x+z}\right)+\dfrac{1}{2}\left(\dfrac{z}{y+z}+\dfrac{x}{x+y}\right)+\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)=\dfrac{3}{2}\)
\(P_{max}=\dfrac{3}{2}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\) hay \(a=b=c=\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\sqrt{22-12\sqrt{2}}+\sqrt{6+4\sqrt{2}}=\sqrt{\left(3\sqrt{2}-2\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}\\ =3\sqrt{2}-2+2+\sqrt{2}=4\sqrt{2}\\ b,\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n}-\sqrt{n+1}}{n-n-1}\\ =\dfrac{\sqrt{n}-\sqrt{n+1}}{-1}=\sqrt{n+1}-\sqrt{n}\)
a) \(\sqrt{22-12\sqrt{2}}+\sqrt{6+4\sqrt{2}}\)
\(=\sqrt{\left(3\sqrt{2}-2\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=3\sqrt{2}-2+2+\sqrt{2}=4\sqrt{2}\)
b) \(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Đặt \(a^2=x;b^2=y;c^2=z\)
Ta có:\(\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}\le\frac{3}{\sqrt{2}}\)
Áp dụng BĐT cô si ta có:
\(\sqrt{\frac{x}{x+y}}=\frac{1}{\sqrt{2}}\sqrt{\frac{4x\left(x+y+z\right)}{3\left(x+y\right)\left(x+z\right)}\frac{3\left(x+z\right)}{2\left(x+y+z\right)}}\)
\(\le\frac{1}{2\sqrt{2}}\left[\frac{4x\left(x+y+z\right)}{3\left(x+y\right)\left(x+z\right)}+\frac{3\left(x+z\right)}{2\left(x+y+z\right)}\right]\)
Tương tự với \(\sqrt{\frac{y}{y+z}}\)và \(\sqrt{\frac{z}{z+x}}\)
Cộng lại ta được:
\(\frac{\sqrt{2}}{3}\left[\frac{x\left(x+y+z\right)}{\left(x+y\right)\left(x+z\right)}+\frac{y\left(x+y+z\right)}{\left(y+z\right)\left(y+x\right)}+\frac{z\left(x+y+z\right)}{\left(z+x\right)\left(z+y\right)}\right]+\frac{3}{2\sqrt{2}}\le\frac{3}{2\sqrt{2}}\)
Sau đó bình phương hai vế rồi
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8xyz\)đẳng thức đúng
Vậy...
Bài 2:
Trước hết ta chứng minh bất đẳng thức sau:
\(\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\le\frac{1}{3}\)
Nhân cả hai vế bđt với 4(a+b+c)4(a+b+c) rồi thu gọn ta được bđt sau:
\(\frac{4a\left(a+b+c\right)}{4a+4b+c}+\frac{4b\left(a+b+c\right)}{4b+4c+a}+\frac{4c\left(a+b+c\right)}{4c+4a+b}\)\(\le\frac{4}{3}\left(a+b+c\right)\)
\(\left[\frac{4a\left(a+b+c\right)}{4a+4b+}-a\right]+\left[\frac{4b\left(a+b+c\right)}{4b+4c+a}-b\right]+\left[\frac{4c\left(a+b+c\right)}{4c+4a+b}-c\right]\le\frac{a+b+c}{3}\)
\(\frac{ca}{4a+4b+c}+\frac{ab}{4b+4c+a}+\frac{bc}{4c+4a+b}\le\frac{a+b+c}{9}\)
Áp dụng bđt cauchy-Schwarz ta có \(\frac{ca}{4a+4b+c}=\frac{ca}{\left(2b+c\right)+2\left(2a+b\right)}\)\(\le\frac{ca}{9}\left(\frac{1}{2b+c}+\frac{2}{2a+b}\right)\)
Từ đó ta có:
\(\text{∑}\frac{ca}{4a+4b+c}\le\frac{1}{9}\text{∑}\left(\frac{ca}{2b+c}+\frac{2ca}{2a+b}\right)\)\(=\frac{1}{9}\left(\text{ ∑}\frac{ca}{2b+c}+\text{ ∑}\frac{2ca}{2a+b}\right)\)\(=\frac{1}{9}\left(\text{ ∑}\frac{ca}{2b+c}+\text{ ∑}\frac{2ab}{2b+c}\right)=\frac{a+b+c}{9}\)
Đặt VT=A rồi áp dụng bđt cauchy-Schwarz cho VT ta có
\(T^2\le3\left(\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\right)\)\(\le3\cdot\frac{1}{3}=1\Leftrightarrow T\le1\)
Dấu = xảy ra khi a=b=c
c bạn tự làm nhé mình mệt rồi :D
![](https://rs.olm.vn/images/avt/0.png?1311)
B đâu ra chỉ? Không biết đề có sai không chứ mình rút gọn ra nhiêu đây thì ko đủ chứng minh C\(\ge0\) được
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
a2 = (1 - a)/(2√2)
=> a4 + a + 1
= (1 - a)^2/8 + a + 1
= (1 - 2a + a^2 + 8a + 8)/8
= (a + 3)^2/8
=> VT = |a + 3|/(2√2) + a^2
Làm nốt