Câu 3 : Ta có :

\(\left\{{}\begin{matrix}a^2+1=a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\\b^2+1=b^2+ab+bc+ca=\left(b+c\right)\left(b+a\right)\\c^2+1=c^2+ab+bc+ca=\left(c+a\right)\left(c+b\right)\end{matrix}\right.\)

Thay vào biểu thức ta được :

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}=\sqrt{\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(b+a\right)\left(c+a\right)\left(c+b\right)}=\sqrt{\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2}=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Vậy biểu thức trên là một số hữu tỉ .

Wish you study well !!!

thank nhiều nhiều nha

Bài 1: 

a: \(=\dfrac{1}{mn^2}\cdot\dfrac{n^2\cdot\left(-m\right)}{\sqrt{5}}=\dfrac{-\sqrt{5}}{5}\)

b: \(=\dfrac{m^2}{\left|2m-3\right|}=\dfrac{m^2}{3-2m}\)

c: \(=\left(\sqrt{a}+1\right):\dfrac{\left(a-1\right)^2}{\left(1-\sqrt{a}\right)}=\dfrac{-\left(a-1\right)}{\left(a-1\right)^2}=\dfrac{-1}{a-1}\)

Lời giải:

Đặt \((\sqrt{a}; \sqrt{b}; \sqrt{c})=(x,y,z)\)

Khi đó điều kiện của bài toán trở thành:

\(x^2+y^2+z^2=x+y+z=2\Rightarrow xy+yz+xz=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\frac{2^2-2}{2}=1\)

Ta có:

\(\frac{\sqrt{a}}{a+1}+\frac{\sqrt{b}}{b+1}+\frac{\sqrt{c}}{c+1}=\frac{x}{x^2+xy+yz+xz}+\frac{y}{y^2+xy+yz+xz}+\frac{z}{z^2+xy+yz+xz}\)

\(=\frac{x}{x(x+y)+z(x+y)}+\frac{y}{y(y+x)+z(y+x)}+\frac{z}{z(z+y)+x(y+z)}\)

\(=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+x)(y+z)}+\frac{z}{(z+x)(z+y)}\)

\(=\frac{x(y+z)+y(x+z)+z(x+y)}{(x+y)(y+z)(x+z)}=\frac{2(xy+yz+xz)}{(x+y)(y+z)(x+z)}=\frac{2}{(x+y)(y+z)(x+z)}(*)\)

Và:

\(\frac{2}{\sqrt{(a+1)(b+1)(c+1)}}=\frac{2}{\sqrt{(x^2+1)(y^2+1)(z^2+1)}}\)

\(=\frac{2}{\sqrt{(x^2+xy+yz+xz)(y^2+xy+yz+xz)(z^2+xy+yz+xz)}}=\frac{2}{\sqrt{(x+y)(x+z)(y+z)(y+x)(z+x)(z+y)}}\)

\(=\frac{2}{\sqrt{(x+y)^2(y+z)^2(z+x)^2}}=\frac{2}{(x+y)(y+z)(x+z)}(**)\)

Từ \((*);(**)\Rightarrow \) đpcm.

Bài 1:

dự đoán dấu = sẽ là \(a^2=b^2=c^2=\dfrac{1}{2}\) nên cứ thế mà chém thôi .

Ta có: \(\left(a^2+1\right)\left(b^2+1\right)=\left(a^2+\dfrac{1}{2}\right)\left(\dfrac{1}{2}+b^2\right)+\dfrac{1}{2}\left(a^2+b^2\right)+\dfrac{3}{4}\)

Bunyakovsky:\(\left(a^2+\dfrac{1}{2}\right)\left(\dfrac{1}{2}+b^2\right)+\dfrac{1}{2}\left(a^2+b^2\right)+\dfrac{3}{4}\ge\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{4}\left(a+b\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\left[\left(a+b\right)^2+1\right]\)

\(VT=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\dfrac{3}{4}\left[\left(a+b\right)^2+1\right]\left(1+c^2\right)\ge\dfrac{3}{4}\left(a+b+c\right)^2\)(đpcm)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{2}}\)

P/s: còn 1 cách khác nữa đó là khai triển sau đó xài schur . Chi tiết trong tệp BĐT schur .pdf

Làm sao có thể dự đoán được dấu "=" trong bài này vậy ạ ?

\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=4\)

\(\Leftrightarrow\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=1\)

\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)

Tương tự: \(b+1=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\)

\(c+1=\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)\)

\(VT=\sum\dfrac{\sqrt{a}}{a+1}=\sum\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)

\(=\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)

\(=\dfrac{2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)

\(VP=\dfrac{2}{\sqrt{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}=\dfrac{2}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2}}\)

\(=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)

\(\Rightarrow VT=VP\) (đpcm)

a) CM:\(\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)

\(\Leftrightarrow n+1+n=\left(n+1-n\right)\left(n+1+n\right)\)

\(\Leftrightarrow2n+1=1\left(2n+1\right)\)

\(\Leftrightarrow2n+1=2n+1\)

\(\Rightarrow\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)

Câu b) ý 2:

Áp dụng BĐT cô si ta có :

\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\\ \dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\\ \dfrac{c}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{c}{b}}\\ \Leftrightarrow2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\ge2\left(\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\right)\\ \Rightarrowđpcm\)