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Cho a,b,c thỏa mãn \(\dfrac{a b-c}{c}=\dfrac{b c-a}{a}=\dfrac{c a-b}{b}\) Tính giá trị M = \(\left(1 \dfrac{b}{a}\right... - Hoc24
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Lời giải:
a, \(A=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+cb}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=1,5\) (AM-GM với a,b,c\(>0\))
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Chú ý: bn cx có thể cm: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\left(a,b,c>0\right)\)để suy ra
b, \(B=\dfrac{a}{b+c}+\dfrac{b+c}{a}+\dfrac{b}{a+c}+\dfrac{a+c}{b}+\dfrac{c}{a+b}+\dfrac{a+b}{c}\)
\(\ge6\sqrt[6]{\dfrac{a}{b+c}.\dfrac{b+c}{a}.\dfrac{b}{a+c}.\dfrac{a+c}{b}.\dfrac{c}{a+b}.\dfrac{a+b}{c}}=6\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Chú ý: bn cx có thể nhóm tổng trên thanh ba nhóm, mỗi nhóm hai hạng tử
a)Đặt \(A=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(A+3=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}\)
\(A+3=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\ge\dfrac{9\left(a+b+c\right)}{2\left(a+b+c\right)}\ge\dfrac{9}{2}\)
\(\Rightarrow A\ge\dfrac{3}{2}\)
\(\Rightarrow MINA=\dfrac{3}{2}\Leftrightarrow a=b=c\)
\(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)
\(\Leftrightarrow a\left(b+c\right)< b\left(a+c\right)\)
\(\Leftrightarrow ab+ac< ba+bc\)
\(\Leftrightarrow ac< bc\)
\(\Leftrightarrow a< b\)(đúng)
a)Áp dụng
\(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}=2\left(1\right)\)
Lại có:\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a}{a+b+c}+\dfrac{b}{b+c+a}+\dfrac{c}{c+a+b}=1\left(2\right)\)
Từ (1) và (2)=> đpcm
Vì \(\dfrac{a}{b}< 1\Rightarrow a< b\Rightarrow ac< bc\Rightarrow ac+ab< bc+ab\Rightarrow a\left(b+c\right)< b\left(a+c\right)\Rightarrow\dfrac{a\left(b+c\right)}{b\left(b+c\right)}< \dfrac{b\left(a+c\right)}{b\left(b+c\right)}\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+c}\)a) ta có
\(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}\)\(\Leftrightarrow\dfrac{a+b+c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{2\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)
\(S=\dfrac{a}{a+b+c}+\dfrac{a+b+c}{a}+\dfrac{b}{a+b+c}+\dfrac{a+b+c}{b}+\dfrac{c}{a+b+c}+\dfrac{a+b+c}{c}-\dfrac{a}{b}-\dfrac{a}{c}-\dfrac{b}{a}-\dfrac{b}{c}-\dfrac{c}{a}-\dfrac{c}{b}=\dfrac{a+b+c}{a+b+c}+\dfrac{a+b+c-b-c}{a}+\dfrac{a+b+c-a-c}{b}+\dfrac{a+b+c-a-b}{c}=1+1+1+1=4\)