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Áp dụng : x + y + z = 0 suy ra x3 + y3 + z3 = 3xyz
1/a + 1/2b + 1/3c = 0 = >... rồi biến đổi nhé
Ta có : \(\frac{a^2-bc}{a}+\frac{b^2-ac}{b}+\frac{c^2-ab}{c}=0\)
=> \(a-\frac{bc}{a}+b-\frac{ac}{b}+c-\frac{ab}{c}=0\)
=> \(a+b+c=\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\)
=> \(a+b+c=abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
=> \(\frac{a+b+c}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> \(\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> \(\frac{2}{bc}+\frac{2}{ac}+\frac{2}{ab}=\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}\)
=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{bc}-\frac{2}{ac}-\frac{2}{ac}=0\)
=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{a^2}-\frac{2}{ac}+\frac{1}{c^2}\right)+\left(\frac{1}{b^2}-\frac{1}{bc}+\frac{1}{c^2}\right)=0\)
=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{a}-\frac{1}{c}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2=0\)
=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{a}-\frac{1}{c}=0\\\frac{1}{b}-\frac{1}{c}=0\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{a}=\frac{1}{b}\\\frac{1}{a}=\frac{1}{c}\\\frac{1}{b}=\frac{1}{c}\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\left(\text{đpcm}\right)\)
1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)
\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)
\(VT=\frac{a^3}{a^2+abc}+\frac{b^3}{b^2+abc}+\frac{c^3}{c^2+abc}\)
Xét \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)
\(\Rightarrow VT=\frac{a^3}{a^2+ab+bc+ac}+\frac{b^3}{b^2+ab+bc+ac}+\frac{c^3}{c^2+ab+bc+ac}\)
\(\Leftrightarrow VT=\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{b^3}{\left(b+a\right)\left(b+c\right)}+\frac{c^3}{\left(c+b\right)\left(c+a\right)}\)
Áp dụng bdt Cauchy ta có :
\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)
Thiết lập tương tự và thu lại ta có :
\(VT+\frac{a+b+c}{2}\ge\frac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\frac{3}{4}\left(a+b+c\right)--\frac{1}{2}\left(a+b+c\right)=\frac{a+b+c}{4}\left(đpcm\right)\)
Dấu " = " xảy ra khi \(a=b=c=3\)
Chúc bạn học tốt !!!
\(VT=\frac{a^3}{a^2+abc}+\frac{b^3}{b^2+abc}+\frac{c^3}{c^2+abc}\)
Xét \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)
\(\Rightarrow VT=\frac{a^3}{a^2+ab+bc+ac}+\frac{b^3}{b^2+ab+bc+ac}+\frac{c^3}{c^2+ab+bc+ac}\)
\(\Leftrightarrow VT=\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{b^3}{\left(b+a\right)\left(b+c\right)}+\frac{c^3}{\left(c+b\right)\left(c+a\right)}\)
Áp dụng BĐT Cauchy ta có :
\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)
Thiết lập tương tự và thu lại ta có :
\(VT+\frac{a+b+c}{2}\ge\frac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{2}\left(a+b+c\right)=\frac{a+b+c}{4}\left(đpcm\right)\)
Dấu " = " xảy ra khi \(a=b=c=3\)
Chúc bạn học tốt !!!
\(A=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\left(a^3+b^3+c^3\right)\frac{1}{abc}\)
Cm với a+b+c=0 thì \(a^3+b^3+c^3=3abc\)(1) .Từ đó tính dc A, muốn cm(1) bạn xét hiệu nhé
\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)(luôn đúng vì a+b+c=0)
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}\\\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}=\frac{1}{c}\\\frac{1}{b}=\frac{1}{a}\end{cases}}\)
\(\Leftrightarrow a=b=c\)
Vậy P =1
day la toan lop 7
ban co biet giai ko ,, giup mk voi