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Bài 1a):
Ta có:
\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\left(a+b\right).\dfrac{a+b}{ab}=\dfrac{a^2+2ab+b^2}{ab}=\dfrac{a^2+b^2}{ab}+2\)
Lại có: (a - b)2 = a2 - 2ab + b2 \(\ge\) 0
\(\Rightarrow\) a2 + b2 \(\ge\) 2ab
\(\Rightarrow\) \(\dfrac{a^2+b^2}{ab}\ge2\)
\(\Rightarrow\) \(\dfrac{a^2+b^2}{ab}+2\ge4\)
Vậy \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)
Bài 2a):
Ta có: \(\left(\sqrt{a}-\sqrt{b}\right)^2=a-2\sqrt{ab}+b\ge0\)
\(\Rightarrow a+b\ge2\sqrt{ab}\)
Vậy ta có đpcm
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
=> \(\dfrac{abc}{ac+bc}=\dfrac{abc}{ab+ac}=\dfrac{abc}{bc+ab}\)
=> ac + bc = ab + ac = bc + ab (do abc \(\ne0\))
=> ac + bc - ab - ac = 0
=> bc - ab = 0
=> b(c - a) = 0
Mà b \(\ne0\) nên c - a = 0 => c = a
Tương tự ta có: a = b
Từ đó có: a = b = c
Thay vào M được:
\(M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
b)\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}\) và \(\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
\(\Rightarrow1+\dfrac{a+b}{c}=1+\dfrac{b+c}{a}\)và \(1+\dfrac{b+c}{a}=1 +\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{c}{c}+\dfrac{a+b}{c}=\dfrac{a}{a}+\dfrac{b+c}{a}\)và \(\dfrac{a}{a}+\dfrac{b+c}{a}=\dfrac{b}{b}+\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{a}\)và \(\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}-\dfrac{a+b+c}{a}=0\) \(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{c}-\dfrac{1}{a}\right)=0\)
và \(\dfrac{a+b+c}{a}-\dfrac{a+b+c}{b}=0\)
\(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{a}-\dfrac{1}{b}\right)=0\)
+) Vì a,b,c đôi một khác 0
\(\Rightarrow a+b+c=0\)
\(\rightarrow a+b=\left(-c\right)\)
\(\rightarrow a+c=\left(-b\right)\)
\(\rightarrow b+c=\left(-a\right)\)
+) Ta có:
\(M=\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{a}\right)\cdot\left(\dfrac{c+a}{c}\right)\)
\(=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}\)
\(=\left(-1\right)\)
Ta có:
\(A=\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}\)
\(A=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}\)
\(A=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\)
Ta lại có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\left(-\dfrac{1}{c}\right)^3\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+3.\dfrac{1}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{1}{c^3}\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=-3.\dfrac{1}{ab}.\dfrac{1}{-c}\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\left(1\right)\)
Thay (1) vào A ta được:
\(A=abc.\dfrac{3}{abc}\)
\(A=3\)