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\(\frac{a}{ab+a+1}=\frac{ac}{abc+ac+c}=\frac{ac}{1+ac+c}\)
\(\frac{b}{bc+b+1}=\frac{abc}{acbc+acb+ac}=\frac{1}{c+1+ac}\)
\(\Leftrightarrow\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}=\frac{ac+1+c}{ac+1+c}=1\)
p/s: cộng lại chỉ = 1 thui >: có sai đề ko vại ?????????
Từ \(abc=1\Rightarrow a=\frac{1}{bc}\) thay vào ta có:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)
\(=\frac{\frac{1}{bc}}{\frac{1}{bc}\cdot b+\frac{1}{bc}+1}+\frac{b}{bc+b+1}+\frac{c}{\frac{1}{bc}\cdot c+c+1}\)
\(=\frac{1}{bc\left(\frac{1}{c}+\frac{1}{bc}+1\right)}+\frac{b}{bc+b+1}+\frac{c}{\frac{1}{b}+c+1}\)
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{b\left(\frac{1}{b}+c+1\right)}\)
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}\)
\(=\frac{1+b+bc}{bc+b+1}=1\)
a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1)
=abc/(ab+a+1)bc+b/(bc+b+1)+bc/(ac+c+1)b
=1/(abcb+abc+bc)+b/(bc+b+1)+bc/(abc+bc+b)
=1/(bc+b+1)+b/(bc+b+1)+bc/(bc+b+1)
=(bc+b+1)/(bc+b+1)=1
Thay abc = 1 vào biểu thức:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}=\frac{a}{a.\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{c}{c.a.\left(b+1+bc\right)}.\)
\(=\frac{a}{a.\left(b+1+bc\right)}+\frac{ba}{a.\left(bc+b+1\right)}+\frac{1}{a.\left(b+1+bc\right)}\)
\(=\frac{ab+a+1}{a.\left(b+1+bc\right)}=\frac{a.\left(b+1+bc\right)}{a.\left(b+1+bc\right)}=1\)
=> đpcm
Đặt \(A=abc\left(bc+a^2\right)\left(ac+b^2\right)\left(ab+c^2\right)\)
Do a; b; c > 0 => A > 0
Giả sử \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a+b}{bc+a^2}-\frac{b+c}{ac+b^2}-\frac{c+a}{ab+c^2}\ge0\)
\(\Leftrightarrow\frac{a^4b^4+b^4c^4+c^4a^4-a^4b^2c^2-b^4a^2c^2-c^4a^2b^2}{A}\ge0\)( tự quy đồng rồi rút gọn nhé, làm chi tiết dài lắm )
\(\Leftrightarrow\frac{2a^4b^4+2b^4c^4+2c^4a^4-2a^4b^2c^2-2b^4a^2c^2-2c^4a^2b^2}{A}\ge0\)
\(\Leftrightarrow\frac{\left(a^2b^2+b^2c^2\right)^2+\left(b^2c^2+c^2a^2\right)^2+\left(c^2a^2+a^2b^2\right)^2}{A}\ge0\)(đúng)
Vậy \(\frac{a+b}{bc+a^2}+\frac{b+c}{ca+b^2}+\frac{c+a}{ab+c^2}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)(đpcm)
\(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow\frac{a^2+b^2+c^2}{abc}\ge\frac{ab+bc+ca}{abc}\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\) ( luôn đúng )
\(\Leftrightarrow\) ĐPCM
Lần lượt áp dụng bất đẳng thức Cô - si có 3 và 4 số, ta có:
\(\frac{a}{18}+\frac{b}{24}+\frac{2}{ab}\ge3.\sqrt[3]{\frac{a}{18}.\frac{b}{24}.\frac{2}{ab}}=\frac{1}{2}\)
\(\frac{a}{9}+\frac{c}{6}+\frac{2}{ac}\ge3.\sqrt[3]{\frac{a}{9}.\frac{c}{6}.\frac{2}{ac}}=1\)
\(\frac{b}{16}+\frac{c}{8}+\frac{2}{bc}\ge3.\sqrt[3]{\frac{b}{16}.\frac{c}{8}.\frac{2}{bc}}=\frac{3}{4}\)
\(\frac{a}{9}+\frac{b}{12}+\frac{c}{6}+\frac{8}{abc}\ge4.\sqrt[4]{\frac{a}{9}.\frac{b}{12}.\frac{c}{6}.\frac{8}{abc}}=\frac{4}{3}\)
\(\frac{13a}{18}+\frac{13b}{24}\ge2\sqrt{\frac{13a}{18}.\frac{13b}{24}}\ge2\sqrt{\frac{13.13.12}{18.24}}=\frac{13}{3}\)
\(\frac{13c}{24}+\frac{13b}{48}\ge2\sqrt{\frac{13c}{24}.\frac{13b}{48}}\ge2\sqrt{\frac{13.13.8}{24.48}}=\frac{13}{6}\)
Cộng vế với vế ta có:
\(a+b+c+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+\frac{8}{abc}\ge\frac{121}{12}\)
\(\frac{b}{bc+b+1}+\frac{a}{ab+a+1}+\frac{c}{ac+c+1}\)
\(=\frac{ac.b}{ac\left(bc+b+1\right)}+\frac{c.a}{c\left(ab+a+1\right)}+\frac{c}{ac+c+1}\)
\(=\frac{1}{c+1+ac}+\frac{ac}{1+ac+c}+\frac{c}{ac+c+1}=1\)
a= b+c=a : b=a+c; c= a=b voi nhung bai nhan chia cung vay