\(a,\sqrt{22-12\sqrt{2}}+\sqrt{6+4\sqrt{2}}=\sqrt{\left(3\sqrt{2}-2\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}\\ =3\sqrt{2}-2+2+\sqrt{2}=4\sqrt{2}\\ b,\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n}-\sqrt{n+1}}{n-n-1}\\ =\dfrac{\sqrt{n}-\sqrt{n+1}}{-1}=\sqrt{n+1}-\sqrt{n}\)

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) \(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)

a.

Bình phương 2 vế, BĐT cần chứng minh trở thành:

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge6\)

Ta có:

\(\sqrt{\left(a^2+1\right)\left(1+b^2\right)}\ge\sqrt{\left(a+b\right)^2}=a+b\)

Tương tự cộng lại:

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge2\left(a+b+c\right)=6\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

b.

\(\sum\dfrac{a+1}{a^2+2a+3}=\sum\dfrac{a+1}{a^2+1+2a+2}\le\sum\dfrac{a+1}{4a+2}\)

Nên ta chỉ cần chứng minh:

\(\sum\dfrac{a+1}{4a+2}\le1\Leftrightarrow\sum\dfrac{4a+4}{4a+2}\le4\)

\(\Leftrightarrow\sum\dfrac{1}{2a+1}\ge1\)

Đúng đo: \(\dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{2c+1}\ge\dfrac{9}{2\left(a+b+c\right)+3}=1\)

Ta có:

VT: \(\left(\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}\right):\dfrac{1}{\sqrt{2}}\)

\(=\left[\dfrac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\right]:\dfrac{1}{\sqrt{2}}\)

\(=\left[\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{\left(\sqrt{3}\right)^2-1^2}\right]:\dfrac{1}{\sqrt{2}}\)

\(=\dfrac{4}{2}:\dfrac{1}{\sqrt{2}}\)

\(=2:\dfrac{1}{\sqrt{2}}\)

\(=2\sqrt{2}\left(dpcm\right)\)

\(VT=\left(\sqrt{3}+1-\sqrt{3}+1\right)\cdot\sqrt{2}=2\cdot\sqrt{2}=VP\)

@Doraemon2611.

:))

\(=\dfrac{\sqrt{3}-1}{\sqrt{2}+1}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{2}{1}=2\)

Đặt A=\(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}\)

⇒A\(^2\)=\(\left(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}\right)^2\)

=\(\sqrt{2}+1+\sqrt{2}-1-2.\sqrt{\sqrt{2}+1}.\sqrt{\sqrt{2}-1}\)

=2.\(\sqrt{2}-\)\(2.\sqrt{2-1}\)

=\(2.\left(\sqrt{2}-1\right)\)

⇒A=\(\sqrt{2\sqrt{2-1}}\)

Vậy \(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)

Áp dụng BĐT cosi cho 2 số dương

\(1=a^2+b^2\ge2ab\Leftrightarrow ab\le\dfrac{1}{2}\)

Mà \(\left(a+b\right)^2=1+2ab\le1+2\cdot\dfrac{1}{2}=2\Leftrightarrow a+b\le\sqrt{2}\)

Áp dụng BĐT Bunhiacopski

\(\left(a\sqrt{1+b}+b\sqrt{1+a}\right)^2\le\left(a^2+b^2\right)\left(1+b+1+a\right)=2+a+b\le2+\sqrt{2}\\ \Leftrightarrow a\sqrt{1+b}+b\sqrt{1+a}\le\sqrt{2+\sqrt{2}}\)

Dấu \("="\Leftrightarrow\dfrac{a}{b}=\sqrt{\dfrac{1+b}{1+a}}\Leftrightarrow a=b=\dfrac{1}{2}\)

Áp dụng BĐT Bunhicopski:

\(\left(a\sqrt{1+b}+b\sqrt{1+a}\right)\le\left(a^2+b^2\right)\left(1+b+1+a\right)=a+b+2\left(1\right)\)

Ta có: \(a^2+b^2\ge2ab\)(BĐT  Cauchy)

\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow\left(a+b\right)^2\le2\Rightarrow a+b\le\sqrt{2}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\left(a\sqrt{1+b}+b\sqrt{1+a}\right)^2\le2+\sqrt{2}\)

\(\Rightarrow a\sqrt{1+b}+b\sqrt{1+a}\le\sqrt{2+\sqrt{2}}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=\dfrac{\sqrt{2}}{2}\)

a.\(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}>\frac{2}{\sqrt{n}+\sqrt{n+1}}=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+1-n}=2\left(\sqrt{n+1}+\sqrt{n}\right)\)

áp dụng công thức cho biểu thức A có A>\(2\left(-\sqrt{2}+\sqrt{26}\right)>7\left(1\right)\)

(so sánh bình phương 2 số sẽ ra nha)

\(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}< \frac{2}{\sqrt{n}+\sqrt{n-1}}=\frac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-n+1}=2\left(\sqrt{n}-\sqrt{n-1}\right)\)

áp dụng công thức cho biểu thức A ta CM được

A<\(2\left(\sqrt{2}-\sqrt{2-1}+\sqrt{3}-\sqrt{3-1}+...+\sqrt{25}-\sqrt{25-1}\right)\)

=\(2\left(-\sqrt{1}+\sqrt{25}\right)=2\left(-1+5\right)=2\cdot4=8\left(2\right)\)

từ (1) và (2) => ĐPCM

b. tương tự câu a ta CM đc BT đã cho=B>\(2\sqrt{51}-2\)> \(5\sqrt{2}\left(1\right)\)

và B<\(2\sqrt{50}=\sqrt{2}\cdot\sqrt{2\cdot50}=10\sqrt{2}\left(2\right)\)

từ (1) và (2)=>ĐPCM

(bạn nhớ phải biến đổi 1 thành 1/\(\sqrt{1}\) trc khi áp dụng công thức nha)

MỜI BẠN THAM KHẢO