Xét : \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(=\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+\frac{2}{abc}.\left(a+b+c\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)(Vì a + b + c = 0)

\(\Rightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\) (đpcm)

Em thử nha, có gì sai bỏ qua ạ.

Đề cho gọn,Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\) thì \(xy+yz+zx=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}=0\) 

Và \(x+y+z=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ca}{abc}=0\)

Ta có: \(VT=\sqrt{x^2+y^2+z^2}=\sqrt{\left(x+y+z\right)^2-2\left(xy+yz+zx\right)}=0\) (1)

Mặt khác,ta có \(VT=\left|x+y+z\right|=0\) (2)

Từ (1) và (2) ta có đpcm

​Dòng cuối phải là

VP=|x+y+z|=0 

đúng không????

uk,thì e viết nhầm -_-"

\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+0.\frac{2}{abc}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\left(a+b+c\right).\frac{2}{abc}}\)

\(=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)

1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)

\(=ac+bc+c^2+ab\)

\(=a\left(b+c\right)+c\left(b+c\right)\)

\(=\left(b+c\right)\left(a+b\right)\)

CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)

\(b+ca=\left(b+c\right)\left(a+b\right)\)

Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)

Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)

CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)

\(\Rightarrow P\le\frac{1}{2}.3\)

\(\Rightarrow P\le\frac{3}{2}\)

Dấu"="xảy ra \(\Leftrightarrow a=b=c\)

Vậy /...

\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)

\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)

Tương tự rồi cộng lại:

\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)

\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)

Dấu "=" xảy ra tại \(a=b=c=1\)

Bài 1 : 

\(P=\sqrt{\frac{ab}{c+ab}}+\sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ca}{b+ca}}\)

\(P=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}+\sqrt{\frac{bc}{a\left(a+b+c\right)+bc}}\)

\(+\sqrt{\frac{ca}{b\left(a+b+c\right)+ca}}\)

\(P=\sqrt{\frac{ab}{ac+bc+c^2+ab}}+\sqrt{\frac{bc}{a^2+ab+ac+bc}}\)

\(+\sqrt{\frac{ca}{ab+b^2+bc+ca}}\)

\(P=\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}\)

Áp dụng bất đẳng thức Cauchy cho 2 bô só thực không âm

\(\Rightarrow\hept{\begin{cases}\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\le\frac{\frac{a}{a+c}+\frac{b}{b+c}}{2}\\\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{b}{a+b}+\frac{c}{a+c}}{2}\\\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}\le\frac{\frac{a}{a+b}+\frac{c}{b+c}}{2}\end{cases}}\)

\(\Rightarrow VT\)

\(\le\frac{\left(\frac{a}{a+c}+\frac{c}{a+c}\right)+\left(\frac{b}{b+c}+\frac{c}{b+c}\right)+\left(\frac{b}{a+b}+\frac{a}{a+b}\right)}{2}\)

\(\Rightarrow VT\le\frac{\frac{a+c}{a+c}+\frac{b+c}{b+c}+\frac{a+b}{a+b}}{2}=\frac{3}{2}\)

\(\Rightarrow P\le\frac{3}{2}\)

Vậy \(P_{max}=\frac{3}{2}\)

Dấu " = " xảy ra khi \(a=b=c=\frac{1}{3}\)

Chúc bạn học tốt !!!

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=0\)

\(\Leftrightarrow a+b+c=0\)

Xét : \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right).\left(b+c\right).\left(c+a\right)=-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\) luôn chia hết cho 3

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)    (do a+b+c = 0)

=>  \(B=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{ \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

=>   đpcm

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{c+b+a}{abc}\right)\)

Mà a+b+c = 0 nên suy ra:

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{0}{abc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

Ta có: (\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\))^{\(^2\)= \(\frac{1}{a^2}\)+\(\frac{1}{b^2}\)+\(\frac{1}{c^2}\)+\(\frac{2}{abc}\)(\(\frac{a+b+c}{abc}\))}

Mà

​A+B+C= 0

nên: VT = VP (đpcm)