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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:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
Ta có \(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\)\(\Rightarrow3\sqrt[3]{a^2b^2c^2}\le3\Leftrightarrow abc\le1\)
\(\Rightarrow\)\(\frac{1}{1+a^2\left(b+c\right)}\le\frac{1}{abc+a^2\left(b+c\right)}\)\(=\frac{1}{a\left(ab+bc+ca\right)}=\frac{1}{3a}\)
\(CMTT\Rightarrow\frac{1}{1+b^2\left(c+a\right)}\le\frac{1}{3b}\)
\(\frac{1}{1+c^2\left(a+b\right)}\le\frac{1}{3c}\)
\(\Rightarrow VT\le\frac{1}{3a}+\frac{1}{3b}+\frac{1}{3c}\)\(=\frac{ab+bc+ca}{3abc}=\frac{1}{abc}\)
Mẫu bài này khó khử ~v
Ta có: \(\frac{1}{a^3\left(b+c\right)}+\frac{a^3\left(b+c\right)}{4}\ge2\sqrt{\frac{1}{a^3\left(b+c\right)}.\frac{a^3\left(b+c\right)}{4}}=2.\frac{1}{2}=1\)
Thiết lập hai BĐT còn lại tương tự và cộng theo vế,ta có:
\(VT+\frac{\left[a^3\left(b+c\right)+b^3\left(a+c\right)+c^3\left(a+b\right)\right]}{4}\ge3\) (*)
Ta sẽ c/m: \(a^3\left(b+c\right)+b^3\left(a+c\right)+c^3\left(a+b\right)\ge6\) (**)
Thật vậy,áp dụng BĐT Cô si,ta có: \(VT_{\left(^∗^∗\right)}\ge2a^2.a\sqrt{bc}+2b^2.b\sqrt{ac}+2c^2.c\sqrt{ab}\)
\(=2a^2\sqrt{abc.a}+2b^2\sqrt{abc.b}+2c^2\sqrt{abc.c}\)
\(=2a^2\sqrt{a}+2b^2\sqrt{b}+2b^2\sqrt{c}\) (***)
Đặt \(\sqrt{a}=t;\sqrt{b}=u;\sqrt{c}=v\).và \(t.u.v=1\)
(***) trở thành: \(2t^5+2u^5+2v^5=2\left(t^5+u^5+v^5\right)\)
Ta có: \(t^5+u^5+v^5+1+1\ge5\sqrt[5]{t^5u^5v^5.1.1}=5\)
Suy ra \(t^5+u^5+v^5\ge5-2=3\)
Suy ra \(2\left(t^5+u^5+v^5\right)\ge2.3=6\) (****)
Kết hợp (**) ; (***) và (****) suy ra \(a^3\left(b+c\right)+b^3\left(a+c\right)+c^3\left(a+b\right)\ge6\)
Thay vào (1) suy ra \(VT+\frac{\left[a^3\left(b+c\right)+b^3\left(a+c\right)+c^3\left(a+b\right)\right]}{4}\ge VT+\frac{6}{4}\ge3\)
Suy ra \(VT\ge\frac{3}{2}^{\left(đpcm\right)}\)
Dấu "=" xảy ra khi a = b = c = 1
Bài dài quá,có gì sai sót mong bạn thông cảm.Vì khi bài dài,mình làm có thể sẽ bị ngược dấu. :v
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 \(x^2+y^2+z^2-xy-yz-zx=0\)
\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
ta có:\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
=\(\frac{\frac{1}{a^2}}{a\left(b+c\right)}+\frac{\frac{1}{b^2}}{b\left(a+c\right)}+\frac{\frac{1}{c^2}}{c\left(a+b\right)}\)
>= \(\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\)(BĐT Svaxo)=\(\frac{\left(\frac{ab+bc+ca}{abc}\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)
>= \(\frac{3\sqrt[3]{a^2b^2c^2}}{2}\left(BĐTAM-GM\right)=\frac{3}{2}\)(đpcm)
dấu = khi a=b=c=1
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=0\)
\(\Leftrightarrow a+b+c=0\)
Xét : \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right).\left(b+c\right).\left(c+a\right)=-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\) luôn chia hết cho 3
sửa đề là chứng minh nó <=1 nha !
ta có \(\frac{2}{\left(a+1\right)^2+b^2+1}=\frac{2}{a^2+b^2+2a+2}\)
mà \(a^2+b^2\ge2ab\Rightarrow\frac{2}{a^2+b^2+2a+2}\le\frac{2}{2ab+2a+2}=\frac{1}{ab+a+1}\)
tương rự, ta có \(...\le\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)
mà từ abc=1, ta có thể chứng minh \(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}=1\)
=>...<=1(ĐPCM)
dấu = xảy ra <=>a=b=c=1
^_^
\(2\left(a^2+b^2+c^2\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2\left(a^2+b^2+c^2\right)+4\frac{ab+bc+ca}{abc}.\)
\(=2\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)\)(vì abc=1)
\(=2\left(a^2+b^2+c^2+2ab+2bc+2ac\right)\)
\(=2\left(a+b+c\right)^2\)
Ta có \(a+b+c\ge3\sqrt[3]{abc}=3\)(bất đẳng thức cô si cho ba số không âm)
Đặt \(a+b+c=x\ge3\)
Dễ thấy : \(2x^2-7x+3=\left(2x-1\right)\left(x-3\right)\ge0\)
Hay \(2\left(a+b+c\right)^2-7\left(a+b+c\right)+3\ge0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge7\left(a+b+c\right)-3\)
Dấu '=' xảy ra khi \(\hept{\begin{cases}a=b=c\\a+b+c=3\end{cases}\Leftrightarrow}a=b=c=1\)
Đặt A = a + b + c .
Áp dụng BĐT Cosi cho 3 số thực dương ta có : \(A\ge3^3\sqrt{abc}=3\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-7\left(a+b+c\right)+3\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+4\cdot\frac{ab+bc+ca}{abc}-7\left(a+b+c\right)+3\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)-7\left(a+b+c\right)+3\)
\(\Leftrightarrow2\left(a+b+c\right)^2-7\left(a+b+c\right)+3\)
\(\Leftrightarrow2A^2-7A+3=\left(2A-1\right)\left(A-3\right)\ge0\)
Dấu "=" xảy ra khi \(a=b=c=1\)