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t nghĩ đề phải bổ sung là a,b,c > 0 nữa.
Bất đẳng thức đã cho tương đương với :
\(\frac{2\left(a^3+b^3+c^3\right)}{abc}-6+\frac{9\left(a+b+c\right)^2}{a^2+b^2+c^2}-27\ge0\)
\(\Leftrightarrow\frac{2\left(a^3+b^3+c^3-3abc\right)}{abc}+\frac{9\left(a^2+b^2+c^2+2ab+2bc+2ac\right)-27\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}\ge0\)
\(\Leftrightarrow\frac{2\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{abc}-\frac{18\left(a^2+b^2+c^2-ab-bc-ac\right)}{a^2+b^2+c^2}\ge0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)\left(\frac{a+b+c}{abc}-\frac{9}{a^2+b^2+c^2}\right)\ge0\)
\(\Leftrightarrow\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\left[\left(a+b+c\right)\left(a^2+b^2+c^2\right)-9abc\right]\ge0\)
cần chứng minh \(\left(a+b+c\right)\left(a^2+b^2+c^2\right)-9abc\ge0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc+a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)-6abc\ge0\)
Ta thấy \(a^3+b^3+c^3-3abc=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)
\(a\left(b^2+c^2\right)+b\left(a^2+c^2\right)+c\left(a^2+b^2\right)-6abc=a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2\ge0\)
Dấu bằng xảy ra khi a = b = c
Thanh Tùng DZ Sao anh ko dùng co si cho nhanh để cm cái bđt cuối ??
\(a+b+c\ge3\sqrt[3]{abc};a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2\right)\ge9abc\Rightarrowđpcm\)
Bài 2:
\(hpt\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+2x\right)+\left(y^2+2y\right)=6\\\left(x^2+2x\right)\left(y^2+2y\right)=9\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2+2x=a\\y^2+2y=b\end{matrix}\right.\) thì:\(\left\{{}\begin{matrix}a+b=6\\ab=9\end{matrix}\right.\)
Từ \(a+b=6\Rightarrow a=6-b\) thay vào \(ab=9\)
\(b\left(6-b\right)=9\Rightarrow-b^2+6b-9=0\)
\(\Rightarrow-\left(b-3\right)^2=0\Rightarrow b-3=0\Rightarrow b=3\)
Lại có: \(a=6-b=6-3=3\)
\(\Rightarrow\left\{{}\begin{matrix}x^2+2x=3\\y^2+2y=3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x+3\right)=0\\\left(y-1\right)\left(y+3\right)=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\\\left[{}\begin{matrix}y=1\\y=-3\end{matrix}\right.\end{matrix}\right.\)
Bài 3:
\(BDT\Leftrightarrow\dfrac{1}{a^2\left(b+c\right)}+\dfrac{1}{b^2\left(c+a\right)}+\dfrac{1}{c^2\left(a+b\right)}\ge\dfrac{3}{2}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{a^2\left(b+c\right)}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{1}{a^2\left(b+c\right)}\cdot\dfrac{b+c}{4}}\)\(=2\sqrt{\dfrac{1}{4a^2}}=\dfrac{1}{a}\)
Tương tự cho 2 BĐT còn lại ta có:
\(\dfrac{1}{b^2\left(c+a\right)}+\dfrac{c+a}{4}\ge\dfrac{1}{b};\dfrac{1}{c^2\left(a+b\right)}+\dfrac{a+b}{4}\ge\dfrac{1}{c}\)
Cộng theo vế 3 BĐT trên ta có:
\(\Rightarrow VT+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow VT+\dfrac{a+b+c}{2}\ge\dfrac{9}{a+b+c}\ge\dfrac{9}{3\sqrt[3]{abc}}\)
\(\Rightarrow VT+\dfrac{3\sqrt[3]{abc}}{2}\ge\dfrac{9}{3\sqrt[3]{abc}}\Rightarrow VT+\dfrac{3}{2}\ge3\left(abc=1\right)\)
\(\Rightarrow VT\ge\dfrac{3}{2}\). Tức là \(\dfrac{1}{a^2\left(b+c\right)}+\dfrac{1}{b^2\left(c+a\right)}+\dfrac{1}{c^2\left(a+b\right)}\ge\dfrac{3}{2}\)
Đẳng thức xảy ra khi \(a=b=c=1\)
Làm cho hoàn thiện luôn nè
1)ĐK:x>0
pt trở thành: x2+1+3x\(\sqrt{\dfrac{x^2+1}{x}}\)=10x
<=>\(\dfrac{x^2+1}{x}\)+3\(\sqrt{\dfrac{x^2+1}{x}}\)=10(*)
đặt y=\(\sqrt{\dfrac{x^2+1}{x}}\)(y>0)
(*)<=>y2+3y-10=0
<=>(y+5)(y-2)=0
<=>\(\left[{}\begin{matrix}y=-5\\y=2\end{matrix}\right.\)
vậy y =2(y>0)
<=>\(\sqrt{\dfrac{x^2+1}{x}}\)=2<=>x2+1=4x
<=>x2-4x+1=0<=>\(\left[{}\begin{matrix}x=\sqrt{3}+2\\x=2-\sqrt{3}\end{matrix}\right.\)
3) điều phải cm<=>\(\dfrac{1}{a^2\left(b+c\right)}+\dfrac{1}{b^2\left(a+c\right)}+\dfrac{1}{c^2\left(a+b\right)}\ge\dfrac{3}{2}\)đặt x=\(\dfrac{1}{a}\);y=\(\dfrac{1}{b}\);z=\(\dfrac{1}{c}\)
P<=>\(\dfrac{x^2yz}{y+z}+\dfrac{xy^2z}{x+z}+\dfrac{xyz^2}{x+y}\)
=\(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\)(xyz=1)
đến đây ta có bất đẳng thức quen thuộc trên
A=\(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\)
A+3=\(\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{x+z}+\dfrac{x+y+z}{x+y}\)
=(x+y+z)(\(\dfrac{1}{y+z}+\dfrac{1}{x+z}+\dfrac{1}{x+y}\))(**)
đặt m=x+y;n=y+z;p=x+z
(**)<=>\(\dfrac{m+n+p}{2}\left(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}\right)\ge\dfrac{9}{2}\)(điều suy ra được từ bất đẳng thức cô-si cho 3 số)
=>A\(\ge\)\(\dfrac{3}{2}\)
=>P\(\ge\)\(\dfrac{3}{2}\)
Lời giải:
Áp dụng BĐT AM-GM:
\(\frac{a^4}{(a+2)(b+2)}+\frac{a+2}{27}+\frac{b+2}{27}+\frac{1}{9}\geq 4\sqrt[4]{\frac{a^4}{27.27.9}}=\frac{4a}{9}\)
\(\frac{b^4}{(b+2)(c+2)}+\frac{b+2}{27}+\frac{c+2}{27}+\frac{1}{9}\geq \frac{4b}{9}\)
\(\frac{c^4}{(c+2)(a+2)}+\frac{c+2}{27}+\frac{a+2}{27}+\frac{1}{9}\geq \frac{4c}{9}\)
Cộng theo vế và rút gọn:
\(\frac{a^4}{(a+2)(b+2)}+\frac{b^4}{(b+2)(c+2)}+\frac{c^4}{(c+2)(a+2)}+\frac{2(a+b+c)}{27}+\frac{7}{9}\geq\frac{4(a+b+c)}{9}\)
\(\frac{a^4}{(a+2)(b+2)}+\frac{b^4}{(b+2)(c+2)}+\frac{c^4}{(c+2)(a+2)}\geq \frac{10(a+b+c)}{27}-\frac{7}{9}=\frac{30}{27}-\frac{7}{9}=\frac{1}{3}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(a+c\right)}+\dfrac{1}{c^3\left(a+b\right)}\)
\(=\dfrac{abc}{a^3\left(b+c\right)}+\dfrac{abc}{b^3\left(a+c\right)}+\dfrac{abc}{c^3\left(a+b\right)}\)
\(=\dfrac{bc}{a^2\left(b+c\right)}+\dfrac{ac}{b^2\left(a+c\right)}+\dfrac{ab}{c^2\left(a+b\right)}\)
\(=\dfrac{b^2c^2}{a^2bc\left(b+c\right)}+\dfrac{a^2c^2}{ab^2c\left(a+c\right)}+\dfrac{a^2b^2}{abc^2\left(a+b\right)}\)
\(Cauchy-Schwarz:\)
\(VT\ge\dfrac{\left(bc+ac+ab\right)^2}{abc\left[a\left(b+c\right)+b\left(a+c\right)+c\left(a+b\right)\right]}\)
\(=\dfrac{\left(bc+ac+ab\right)^2}{2\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2}\)
\(AM-GM:\)
\(ab+bc+ca\ge\sqrt[3]{\left(abc\right)^2}=3\)
\(\Rightarrow VT\ge\dfrac{ab+bc+ca}{2}\ge\dfrac{3}{2}\)
\("="\Leftrightarrow a=b=c=1\)
Lời giải khác:
Áp dụng BĐT AM-GM:
\(\frac{1}{a^3(b+c)}+\frac{a(b+c)}{4}\geq 2\sqrt{\frac{1}{4a^2}}=\frac{1}{a}=\frac{abc}{a}=bc\)
\(\frac{1}{b^3(a+c)}+\frac{b(a+c)}{4}\geq 2\sqrt{\frac{1}{4b^2}}=\frac{1}{b}=\frac{abc}{b}=ac\)
\(\frac{1}{c^3(a+b)}+\frac{c(a+b)}{4}\geq 2\sqrt{\frac{1}{4c^2}}=\frac{1}{c}=\frac{abc}{c}=ab\)
Cộng theo vế và rút gọn:
\(\Rightarrow \frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}+\frac{ab+bc+ac}{2}\ge ab+bc+ac\)
\(\Rightarrow \frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\geq \frac{ab+bc+ac}{2}\geq \frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}\) (AM_GM)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
T đề nghị ban EDOGAWA CONAN không dùng nick k\này hỏi rồi lấy nick chính trả lời và tự tick nữa. T biết hai cậu là 1 mà không muốn nói thôi.
P/s:Nếu thế nữa t sẽ báo phynit.
Đặt : \(x=\dfrac{a+b}{a-b}\) ; \(y=\dfrac{b+c}{b-c}\) ; \(z=\dfrac{c+a}{c-a}\)
Ta có : \(\left(x+1\right)\left(y+1\right)\left(z+1\right)=\left(x-1\right)\left(y-1\right)\left(z-1\right)\)
\(\Leftrightarrow xy+yz+zx=-1\)
Mà \(\left(x+y+z\right)^2\ge0\)
\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)\ge0\)
\(\Leftrightarrow x^2+y^2+z^2\ge2\)
\(\Rightarrow\dfrac{\left(a+b\right)^2}{\left(a-b\right)^2}+\dfrac{\left(b+c\right)^2}{\left(b-c\right)^2}+\dfrac{\left(c+a\right)^2}{\left(c-a\right)^2}\ge2\left(đpcm\right)\)
\(BDT\Leftrightarrow2\left[\dfrac{a^3+b^3+c^2}{abc}-3\right]+9\left[\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2}-3\right]\ge0\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)\sum\left(a-b\right)^2}{abc}+\dfrac{-9\sum\left(a-b\right)^2}{a^2+b^2+c^2}\ge0\)
\(\Leftrightarrow\sum\left(a-b\right)^2\left(\dfrac{a+b+c}{abc}-\dfrac{9}{a^2+b^2+c^2}\right)\ge0\)
\(\Leftrightarrow\sum\left(a-b\right)^2.\dfrac{\sum\left(a-b\right)^2.\left(a+b+3c\right)}{2abc\left(a^2+b^2+c^2\right)}\ge0\) (đúng)