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

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

\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=\dfrac{1+1+1}{a+b+c}=\dfrac{3}{a+b+c}=\dfrac{3}{1}=3\)

\(\Rightarrow a=b=c=\dfrac{1}{3}\)

\(\Rightarrow A=\dfrac{a^3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=a^3=\left(\dfrac{1}{3}\right)^3=\dfrac{1}{27}\)

khó thế

sai de roi

với a,b,c khác 0 ta có:\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)  => \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)=>\(\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)  =>\(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)  =>a=b=c  => M=1

Từ giả thiết suy ra \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)(vì a,b,c khác 0)

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

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

\(\Rightarrow M=1\)

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

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

Từ \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\) suy ra \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)

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

\(\Rightarrow\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}\\\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\\\frac{1}{c}+\frac{1}{a}=\frac{1}{a}+\frac{1}{b}\end{cases}}\)\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)

Khi đó \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)

\(P\le a^2+b^2+c^2+3\sqrt{3\left(a^2+b^2+c^2\right)}=12\)

\(P_{max}=12\) khi \(a=b=c=1\)

Lại có: \(\left(a+b+c\right)^2=3+2\left(ab+bc+ca\right)\ge3\Rightarrow a+b+c\ge\sqrt{3}\)

\(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)

\(\Rightarrow\sqrt{3}\le a+b+c\le3\)

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

\(P=\dfrac{1}{2}\left(a+b+c\right)^2+3\left(a+b+c\right)-\dfrac{3}{2}\)

Đặt \(a+b+c=x\Rightarrow\sqrt{3}\le x\le3\)

\(P=\dfrac{1}{2}x^2+3x-\dfrac{3}{2}=\dfrac{1}{2}\left(x-\sqrt{3}\right)\left(x+6+\sqrt{3}\right)+3\sqrt{3}\ge3\sqrt{3}\)

\(P_{min}=3\sqrt{3}\) khi \(x=\sqrt{3}\) hay \(\left(a;b;c\right)=\left(0;0;\sqrt{3}\right)\) và hoán vị

thế bạn bt hok

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) ; \(\forall a;b;c\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Rightarrow ab+bc+ca\le1\)

\(\Rightarrow P_{max}=1\) khi \(a=b=c\)

Lại có:

\(\left(a+b+c\right)^2\ge0\) ; \(\forall a;b;c\)

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

\(\Leftrightarrow ab+bc+ca\ge-\dfrac{a^2+b^2+c^2}{2}=-\dfrac{1}{2}\)

\(P_{min}=-\dfrac{1}{2}\) khi \(a+b+c=0\)

Câu hỏi của Đậu Đình Kiên - Toán lớp 7 - Học toán với OnlineMath