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\(6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)
\(=6\left(x^2+y^2+z^2\right)+12\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)-2\left(xy+yz+xz\right)\)
\(=6\left(x+y+z\right)^2+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{2z+x+y}\right)-2\left(xy+yz+xz\right)\)
\(\ge6\left(x+y+z\right)^2+2.\dfrac{\left(1+1+1\right)^2}{2x+y+z+x+2y+z+2z+x+y}-2\left(xy+yz+xz\right)\)
\(=6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-2\left(xy+yz+xz\right)\)
\(\ge6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-\dfrac{2}{3}\left(x+y+z\right)^2\)
\(=6.\left(\dfrac{3}{4}\right)^2+\dfrac{18}{4.\dfrac{3}{4}}-\dfrac{2}{3}.\left(\dfrac{3}{4}\right)^2=9\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)
a) ab+bc+ca\(\le\dfrac{\left(a+c+b\right)^2}{3}\)
\(\Leftrightarrow3ab+3bc+3ac\le a^2+b^2+c^2+2ab+2bc+2ac\)
\(\Leftrightarrow ab+bc+ac\le a^2+b^2+c^2\)
\(\Leftrightarrow2ab+2bc+2ca\le2a^2+2b^2+2c^2\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng \(\forall a,b,c\)
Lời giải:
Từ \(xy+yz+xz=xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \((a,b,c)=\left(\frac{1}{x}; \frac{1}{y}; \frac{1}{z}\right)\Rightarrow a+b+c=1\)
BĐT cần chứng minh trở thành:
\(P=\frac{c^3}{(a+1)(b+1)}+\frac{a^3}{(b+1)(c+1)}+\frac{b^3}{(c+1)(a+1)}\geq \frac{1}{16}(*)\)
Thật vậy, áp dụng BĐT Cauchy ta có:
\(\frac{c^3}{(a+1)(b+1)}+\frac{a+1}{64}+\frac{b+1}{64}\geq 3\sqrt[3]{\frac{c^3}{64^2}}=\frac{3c}{16}\)
\(\frac{a^3}{(b+1)(c+1)}+\frac{b+1}{64}+\frac{c+1}{64}\geq 3\sqrt[3]{\frac{a^3}{64^2}}=\frac{3a}{16}\)
\(\frac{b^3}{(c+1)(a+1)}+\frac{c+1}{64}+\frac{a+1}{64}\geq 3\sqrt[3]{\frac{b^3}{64^2}}=\frac{3b}{16}\)
Cộng theo vế các BĐT trên và rút gọn :
\(\Rightarrow P+\frac{a+b+c+3}{32}\geq \frac{3(a+b+c)}{16}\)
\(\Leftrightarrow P+\frac{4}{32}\geq \frac{3}{16}\Leftrightarrow P\geq \frac{1}{16}\)
Vậy \((*)\) được chứng minh. Bài toán hoàn tất.
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\Leftrightarrow x=y=z=3\)
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\(\ge\)\(\dfrac{3}{2}\)
3a)\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{2y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{2y-1}=1\end{matrix}\right.\) (ĐK: x≠2;y≠\(\dfrac{1}{2}\))
Đặt \(\dfrac{1}{x-2}=a;\dfrac{1}{2y-1}=b\) (ĐK: a>0; b>0)
Hệ phương trình đã cho trở thành
\(\left\{{}\begin{matrix}a+b=2\\2a-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\2\left(2-b\right)-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\4-2b-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\b=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{7}{5}\left(TM\text{Đ}K\right)\\b=\dfrac{3}{5}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Khi đó \(\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{7}{5}\\\dfrac{1}{2y-1}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\left(x-2\right)=5\\3\left(2y-1\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x-14=5\\6y-3=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{7}\left(TM\text{Đ}K\right)\\y=\dfrac{4}{3}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Vậy hệ phương trình đã cho có nghiệm duy nhất (x;y)=\(\left(\dfrac{19}{7};\dfrac{4}{3}\right)\)
b) Bạn làm tương tự như câu a kết quả là (x;y)=\(\left(\dfrac{12}{5};\dfrac{-14}{5}\right)\)
c)\(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\)(ĐK: x≥1;y≥0)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+4\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49\left(x-1\right)=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49x-49=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{218}{49}\\y=\dfrac{4}{49}\end{matrix}\right.\left(TM\text{Đ}K\right)\)
Bài 4:
Theo đề, ta có hệ:
\(\left\{{}\begin{matrix}3\left(3a-2\right)-2\left(2b+1\right)=30\\3\left(a+2\right)+2\left(3b-1\right)=-20\end{matrix}\right.\)
=>9a-6-4b-2=30 và 3a+6+6b-2=-20
=>9a-4b=38 và 3a+6b=-20+2-6=-24
=>a=2; b=-5
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 \(\left(x+y+z\right)^2\ge0\)
\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)\ge0\)
\(\Leftrightarrow x^2+y^2+z^2\ge2\)
\(\Rightarrow\dfrac{\left(a+b\right)^2}{\left(a-b\right)^2}+\dfrac{\left(b+c\right)^2}{\left(b-c\right)^2}+\dfrac{\left(c+a\right)^2}{\left(c-a\right)^2}\ge2\left(đpcm\right)\)
Bài 1:
\(P=(x+1)\left(1+\frac{1}{y}\right)+(y+1)\left(1+\frac{1}{x}\right)\)
\(=2+x+y+\frac{x}{y}+\frac{y}{x}+\frac{1}{x}+\frac{1}{y}\)
Áp dụng BĐT Cô-si:
\(\frac{x}{y}+\frac{y}{x}\geq 2\)
\(x+\frac{1}{2x}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
\(y+\frac{1}{2y}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
Áp dụng BĐT SVac-xơ kết hợp với Cô-si:
\(\frac{1}{2x}+\frac{1}{2y}\geq \frac{4}{2x+2y}=\frac{2}{x+y}\geq \frac{2}{\sqrt{2(x^2+y^2)}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Cộng các BĐT trên :
\(\Rightarrow P\geq 2+2+\sqrt{2}+\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)
Vậy \(P_{\min}=4+3\sqrt{2}\Leftrightarrow a=b=\frac{1}{\sqrt{2}}\)
Bài 2:
Áp dụng BĐT Svac-xơ:
\(\frac{1}{a+3b}+\frac{1}{b+a+2c}\geq \frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)
\(\frac{1}{b+3c}+\frac{1}{b+c+2a}\geq \frac{4}{2b+4c+2a}=\frac{2}{b+2c+a}\)
\(\frac{1}{c+3a}+\frac{1}{c+a+2b}\geq \frac{4}{2c+4a+2b}=\frac{2}{c+2a+b}\)
Cộng theo vế và rút gọn :
\(\Rightarrow \frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\geq \frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c$