Từ \(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)

\(\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)

\(\Rightarrow\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{b}{bc}+\dfrac{c}{bc}=\dfrac{c}{ca}+\dfrac{a}{ca}\)

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{b}+\dfrac{1}{a}=\dfrac{1}{c}+\dfrac{1}{b}\\\dfrac{1}{c}+\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\\\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{1}{b}+\dfrac{1}{a}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\\\dfrac{1}{c}=\dfrac{1}{b}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\Rightarrow a=b=c\)

Khi đó: \(M=\dfrac{ab+bc+ca}{a^2+b^2+c^2}=\dfrac{1\cdot1+1\cdot1+1\cdot1}{1^2+1^2+1^2}=\dfrac{3}{3}=1\)

Câu hỏi của Conan Kudo - Toán lớp 8 - Học toán với OnlineMath

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

\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+a^2+b^2+2c^2}=\dfrac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)

Tương tự và cộng lại:

\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)

1, Áp dụng BĐT cosi cho a,b,c>0

\(ab+bc\ge2\sqrt{ab^2c}=2b\sqrt{ac}\\ bc+ca\ge2\sqrt{abc^2}=2c\sqrt{ab}\\ ca+ab\ge2\sqrt{a^2bc}=2a\sqrt{bc}\)

Cộng VTV 3 BĐT trên:

\(\Leftrightarrow2\left(ab+bc+ac\right)\ge2\left(b\sqrt{ac}+a\sqrt{bc}+c\sqrt{ab}\right)\\ \Leftrightarrow ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)

\(2,\)

Ta có

 \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\\ \Leftrightarrow a^2+b^2+c^2-ab-ac-bc\ge0\\ \Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

Áp dụng BĐT cm ở câu 1

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