1+1.2=?

cho mình hỏi đề đúng không vậy

\(a^2+\frac{1}{a^2}\ge2\sqrt{a^2+\frac{1}{a^2}}=2\\ \)(do Bđt cosi)=> \(a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge6\\ \)

Dấu "=" xảy ra <=> a=b=c=1

=>B=3

Bất đẳng thức cosi mình chưa học

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

Dấu = xảy ra khi a^4=b^4=c^4=1 <=> \(a=\pm1;b=\pm1;c\pm1\)

-> B = 3

\(a+b+c=2020\Rightarrow\frac{1}{a+b+c}=\frac{1}{2020}\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{bc+ac+ab}{abc}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)-abc=0\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(b+c\right)+a\left(ab+ac\right)+abc-abc=0\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(b+c\right)+a^2\left(b+c\right)=0\)

\(\Leftrightarrow\left(ab+bc+ac+a^2\right)\left(b+c\right)=0\)

\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

Nếu a + b = 0 thì c = 2020

Nếu b + c = 0 thì a = 2020

Nếu a + c = 0 thì b = 2020

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2020}\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Rightarrow\frac{bc+ac+ab}{abc}=\frac{1}{a+b+c}\)

\(\Rightarrow\left(a+b+c\right)\left(ab+ac+bc\right)=abc\)

\(\Rightarrow a^2b+a^2c+abc+ab^2+abc+b^2c+abc+ac^2+bc^2=abc\)

\(\Rightarrow...\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(TH1:a=-b\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a}-\frac{1}{a}+\frac{1}{c}=\frac{1}{c}\)

Mà \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2020}\Rightarrow\frac{1}{c}=\frac{1}{2020}\Leftrightarrow c=2020\)

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