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\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{c+b+a}{abc}\right)\)
Mà a+b+c = 0 nên suy ra:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{0}{abc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
Ta có: (\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\))\(^2\)= \(\frac{1}{a^2}\)+\(\frac{1}{b^2}\)+\(\frac{1}{c^2}\)+\(\frac{2}{abc}\)(\(\frac{a+b+c}{abc}\))
Mà
A+B+C= 0
nên: VT = VP (đpcm)
Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)
Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)
Ta có:
\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)
Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)
\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)
Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Ta có:\(a+b+c=0\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
Xét \(\frac{2a+bc}{a+c}=\frac{a\left(a+b+c\right)+bc}{a+c}=\frac{a^2+ab+ac+bc}{a+c}=\frac{\left(a+b\right)\left(a+c\right)}{a+c}=a+b\)(thay 2=a+b+c)
Tương tự \(\frac{2b+ac}{a+b}=b+c\)và \(\frac{2c+ab}{c+b}=c+a\)
\(\Rightarrow M=\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\)
\(M=2.\left(a+b+c\right)\)
\(M=4\)
Ta có:
\(\left(a+b-c\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge2ac+2bc-2ab\)
Mà \(a^2+b^2+c^2=\frac{5}{3}< 2\)
\(\Rightarrow2ac+2bc-2ab< 2\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}\)
\(a^3+b^3+c^3=3abc\)
<=> \(a^3+b^3+c^3-3abc=0\)
<=> \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
<=> \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)
<=> \(\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
đến đây ez tự làm nốt nhé, ko ra ib mk
"Chấm" nhẹ hóng cao nhân ạ :)
P/s: mong các bác giải theo cách lớp 8 ạ :) Tặng 5SP / 1 câu nhé ;)
cho a,b,c>0 thỏa mãn a+b+c=1
tìm GTLN của P=\(\frac{ab}{c+1}\)+\(\frac{bc}{a+1}\)+\(\frac{ac}{b+1}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{ab}{c+1}=\frac{ab}{\left(c+a\right)+\left(b+c\right)}\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)
Tương tự cho 2 BĐT còn lại ta có:
\(\frac{bc}{a+1}\le\frac{1}{4}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right);\frac{ac}{b+1}\le\frac{1}{4}\left(\frac{ac}{a+b}+\frac{ac}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(P\le\frac{1}{4}\left[\left(\frac{ab}{b+c}+\frac{ac}{b+c}\right)+\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)+\left(\frac{bc}{a+b}+\frac{ac}{a+b}\right)\right]\)
\(=\frac{1}{4}\left[\frac{a\left(b+c\right)}{b+c}+\frac{b\left(a+c\right)}{a+c}+\frac{c\left(a+b\right)}{a+b}\right]\)
\(=\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\cdot1=\frac{1}{4}\left(a+b+c=1\right)\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)