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Có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\) (NHân cả hai vế vs a+b+c)
\(\Leftrightarrow\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{c+a}+\frac{c\left(a+b+c\right)}{a+b}=a+b+c\)
\(\Leftrightarrow\frac{a^2+a\left(b+c\right)}{b+c}+\frac{b^2+b\left(c+a\right)}{c+a}+\frac{c^2+c\left(a+b\right)}{a+b}=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
=> đpcm
Anh/chị tham khảo ở đây nhé :
đặt x=a^2 + 2bc, y=b^2 + 2ac, z=c^2 + 2ab
=> x + y + z = (a + b + c)^2 <(=) 1
VT bpt : 1/x + 1/y + 1/z >(=) 3.căn3(1/xyz)...dùng cô-si cho 3 số nhé
mà x + y + z >(=) 3.căn3(xyz) <(=) 1
<=> 1/( 3.căn3 (xyz) >(=) 1 (ở đây là đổi chiều bđt)
<=> 1/ căn3 (xyz) >(=) 3
=> VT: 1/x + 1/y + 1/z >(=) 3.3 = 9
Áp dụng BĐT AM-GM ta có:
\(\frac{a}{b^2}+\frac{1}{a}\ge2\sqrt{\frac{a}{b^2}\cdot\frac{1}{a}}=2\sqrt{\frac{1}{b^2}}=\frac{2}{b}\)
\(\frac{b}{c^2}+\frac{1}{b}\ge2\sqrt{\frac{b}{c^2}\cdot\frac{1}{b}}=\frac{2}{c}\)
\(\frac{c}{a^2}+\frac{1}{c}\ge2\sqrt{\frac{c}{a^2}\cdot\frac{1}{c}}=\frac{2}{a}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\Leftrightarrow VT\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Dấu "=" xảy ra khi \(a=b=c\)
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)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
\(\Rightarrow2+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
\(\Rightarrow2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2\)
\(\Rightarrow\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=1\)
\(\Rightarrow\frac{a+b+c}{abc}=1\Rightarrow a+b+c=abc\left(đpcm\right)\)
Có: \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)
\(\Leftrightarrow\frac{ayz+bxz+cxy}{xyz}=0\)
\(\Leftrightarrow ayz+bxz+cxy=0\)
Lại có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1-2\cdot\frac{ayz+bxz+cxy}{abc}=1-2\cdot\frac{0}{abc}=1\)
=>đpcm
Nhân \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)với a+b+c