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Ta có \(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\)\(\Rightarrow3\sqrt[3]{a^2b^2c^2}\le3\Leftrightarrow abc\le1\)
\(\Rightarrow\)\(\frac{1}{1+a^2\left(b+c\right)}\le\frac{1}{abc+a^2\left(b+c\right)}\)\(=\frac{1}{a\left(ab+bc+ca\right)}=\frac{1}{3a}\)
\(CMTT\Rightarrow\frac{1}{1+b^2\left(c+a\right)}\le\frac{1}{3b}\)
\(\frac{1}{1+c^2\left(a+b\right)}\le\frac{1}{3c}\)
\(\Rightarrow VT\le\frac{1}{3a}+\frac{1}{3b}+\frac{1}{3c}\)\(=\frac{ab+bc+ca}{3abc}=\frac{1}{abc}\)
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:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
\(1+a^2=a^2+ab+bc+ca=\left(a+b\right)\left(c+a\right)\)
Tương tự, ta có: \(1+b^2=\left(a+b\right)\left(b+c\right)\)\(;\)\(1+c^2=\left(b+c\right)\left(c+a\right)\)
\(\Rightarrow\)\(\frac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) ( do a, b, c dương )
\(\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}=\frac{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
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đây là 1 sự nhầm lẫn đối với các bạn nhác tìm dấu = :))
Sử dụng BĐT Svacxo ta có :
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}\)
\(=\frac{1}{a^2+b^2+c^2}+\frac{18}{2ab+2bc+2ca}\ge\frac{\left(1+\sqrt{18}\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}\)
\(=\frac{19+\sqrt{72}}{\left(a+b+c\right)^2}=\frac{25\sqrt{2}}{1}=25\sqrt{2}\)
bài làm của e :
Áp dụng BĐT Svacxo ta có :
\(Q\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{7}{ab+bc+ca}\)
Theo hệ quả của AM-GM thì : \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=\frac{1}{3}\)
\(< =>\frac{7}{ab+bc+ca}\ge\frac{7}{\frac{1}{3}}=21\)
Tiếp tục sử dụng Svacxo thì ta được :
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{7}{ab+bc+ca}\ge\frac{9}{\left(a+b+c\right)^2}+21=30\)
Vậy \(Min_P=30\)đạt được khi \(a=b=c=\frac{1}{3}\)
Và đương nhiên cách bạn dcv_new chỉ đúng với \(k\ge2\) ở bài:
https://olm.vn/hoi-dap/detail/259605114604.html
Thực ra bài Min \(\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}\) khi a + b + c = 1
chỉ là hệ quả của bài \(\frac{1}{a^2+b^2+c^2}+\frac{k}{ab+bc+ca}\) khi \(a+b+c\le1\)
Ngoài ra nếu \(k< 2\) thì min là: \(\left(1+\sqrt{2k}\right)^2\)
1/ Đặt
\(\frac{a}{b^2}=x,\frac{b}{c^2}=y,\frac{c}{a^2}=z,xyz=1\)thì ta có
\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow xy+yz+zx=x+y+z\)
\(\Leftrightarrow xyz-xy-yz-zx+x+y+z-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)
\(\Leftrightarrow x=1;y=1;z=1\)
\(\Rightarrow\frac{a}{b^2}=1;\frac{b}{c^2}=1;\frac{c}{a^2}=1\)
\(\Leftrightarrow a=b^2;b=c^2;c=a^2\)
2/ Đặt
\(ab=x,bc=y,ca=z\) cần tính
\(P=\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)\left(1+\frac{y}{x}\right)\)
\(\Rightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{cases}}\)
Xét \(x+y+z=0\)
\(\Rightarrow P=\frac{x+y}{x}.\frac{y+z}{y}.\frac{z+x}{z}=\frac{\left(-x\right)\left(-y\right)\left(-z\right)}{xyz}=-1\)
Xét \(x^2+y^2+z^2-xy-yz-zx=0\)
\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
1. Áp dụng BĐT Cauchy dạng Engle, ta có :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)\left(\frac{9}{a+b+c}\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
\(\frac{1}{3}\left(a^3+b^3+a+b\right)+ab\le a^2+b^2+1\)
\(\Leftrightarrow\frac{1}{3}\left(a+b\right)\left(a^2+b^2+1-ab\right)+ab\le a^2+b^2+1\)
\(\Leftrightarrow\left(a^2+b^2+1\right)\left(\frac{a+b}{3}-1\right)-ab\left(\frac{a+b}{3}-1\right)\le0\)
\(\Leftrightarrow\left(a^2+b^2+1-ab\right)\left(\frac{a+b}{3}-1\right)\le0\)
Vì a, b dương \(\Rightarrow a^2+b^2+1-ab>0\Rightarrow\left(\frac{a+b}{3}-1\right)\le0\Leftrightarrow a+b\le3\)
\(M=\frac{a^2+8}{a}+\frac{b^2+2}{b}=a+\frac{8}{a}+b+\frac{2}{b}=2a+2b+\frac{8}{a}+\frac{2}{b}-\left(a+b\right)\ge8+4-3=9\)
Áp dụng BĐT Cauchy cho a ; b dương
Dấu "=" xảy ra \(\Leftrightarrow a=2;b=1\)