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Xét: \(1+c^2=ab+bc+ca+c^2=\left(a+c\right)\left(b+c\right)\)
Tương tự CM được:
\(1+b^2=\left(a+b\right)\left(c+b\right)\) và \(1+a^2=\left(c+a\right)\left(b+a\right)\)
Mặt khác ta tách: \(\hept{\begin{cases}a-b=\left(a+c\right)-\left(b+c\right)\\b-c=\left(a+b\right)-\left(c+a\right)\\c-a=\left(c+b\right)-\left(a+b\right)\end{cases}}\)
Thay vào ta được:
\(Vt=\frac{\left(a+c\right)-\left(b+c\right)}{\left(a+c\right)\left(b+c\right)}+\frac{\left(a+b\right)-\left(c+a\right)}{\left(a+b\right)\left(c+a\right)}+\frac{\left(c+b\right)-\left(a+b\right)}{\left(b+c\right)\left(a+b\right)}\)
\(=\frac{1}{b+c}-\frac{1}{c+a}+\frac{1}{c+a}-\frac{1}{a+b}+\frac{1}{a+b}-\frac{1}{b+c}\)
\(=0\)
=> 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)
Câu 1:
- Chứng minh a3+b3+c3=3abc thì a+b+c=0
\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(\Rightarrow\left[\left(a+b\right)^3+c^3\right]-3abc\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow0=0\) Đúng (Đpcm)
- Chứng minh a3+b3+c3=3abc thì a=b=c
Áp dụng Bđt Cô si 3 số ta có:
\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)
Dấu = khi a=b=c (Đpcm)
Câu 2
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3\cdot\frac{1}{abc}\)
Ta có:
\(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\cdot3\cdot\frac{1}{abc}=3\)
Ta có: \(\left(x+y\right)^2\ge4xy\)
\(\Rightarrow\frac{xy}{x+y}\le\frac{1}{4}\left(x+y\right)\)
\(\Rightarrow\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{1}{4}\left(a+b\right)+\frac{1}{4}\left(b+c\right)+\frac{1}{4}\left(c+a\right)\)
\(=\frac{a+b+c}{2}\)
Dấu \("="\) xảy ra \(\Leftrightarrow a=b=c\)
\(/(/frac//\)la j zay
Áp dụng BĐT Cô si với a,b,c>0 ta có:
\(\frac{bc}{a}+\frac{ca}{b}\ge2\sqrt{\frac{bc}{a}.\frac{ca}{b}}=2\sqrt{c^2}=2c\)
Tương tự \(\frac{ca}{b}+\frac{ab}{c}\ge2a\)
\(\frac{ab}{c}+\frac{bc}{a}\ge2b\)
\(\Rightarrow2\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\ge2\left(a+b+c\right)\)
\(\Rightarrow\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c\)