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Theo đề ra, ta có:
\(a^2+b^2+c^2\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
\(=a^3+b^3+c^3+a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\)
Theo BĐT Cô-si:
\(\left\{{}\begin{matrix}a^3+ab^2\ge2a^2b\\b^3+bc^2\ge2b^2c\\c^3+ca^2\ge2c^2a\end{matrix}\right.\Rightarrow a^2+b^2+c^2\ge3\left(a^2b+b^2c+c^2a\right)\)
Do vậy \(M\ge14\left(a^2+b^2+c^2\right)+\dfrac{3\left(ab+bc+ac\right)}{a^2+b^2+c^2}\)
Ta đặt \(a^2+b^2+c^2=k\)
Luôn có \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=1\)
Vì thế nên \(k\ge\dfrac{1}{3}\)
Khi đấy:
\(M\ge14k+\dfrac{3\left(1-k\right)}{2k}=\dfrac{k}{2}+\dfrac{27k}{2}+\dfrac{3}{2k}-\dfrac{3}{2}\ge\dfrac{1}{3}.\dfrac{1}{2}+2\sqrt{\dfrac{27k}{2}.\dfrac{3}{2k}}-\dfrac{3}{2}=\dfrac{23}{3}\)
\(\Rightarrow Min_M=\dfrac{23}{3}\Leftrightarrow a=b=c=\dfrac{1}{3}\).
Bài này ta dùng bđt Cauchy-Schwaz
VT=\(\frac{\left(bc\right)^2}{a^2bc\left(b+c\right)}\)\(+\frac{\left(\text{c}\text{a}\right)^2}{\text{b}^2c\text{a}\left(\text{c}+\text{a}\right)}\)\(+\frac{\left(\text{a}\text{b}\right)^2}{\text{c}^2\text{a}\text{b}\left(\text{a}+b\right)}\)
\(\ge\)\(\frac{\left(ab+bc+ca\right)^2}{2abc\left(ab+bc+ca\right)}\)\(=\frac{ab+bc+ca}{2abc}\)\(=\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\)\(=\)VP
=> đpcm
Dấu \("="\)xảy ra <=> a=b=c
1/\(=4a^2+4b^2+c^2+8ab-4bc-4ca+4b^2+4c^2+a^2+8bc-4ca-4ab+4a^2+4c^2+b^2+8ca-4bc-4ab=\)
\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)
2/
Ta có
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge-2\left(ab+bc+ca\right)=2\)
\(\Rightarrow P=9\left(a^2+b^2+c^2\right)\ge18\)
\(\Rightarrow P_{min}=18\)
a)
$3^x-y^3=1$
$\Leftrightarrow 3^x=y^3+1$
$\Leftrightarrow 3^x=(y+1)(y^2-y+1)$
$\gcd(y+1,y^2-y+1)=\gcd(y+1,3)$
Vì $3^x$ chỉ có ước nguyên tố là $3$ nên
$y+1=3^m,\quad y^2-y+1=3^n\qquad (m,n\in\mathbb N,\ m+n=x)$
Ta có $y^2-y+1-(y+1)(y-2)=3$ nên $\gcd(y+1,y^2-y+1)\mid 3$
Suy ra $\gcd(y+1,y^2-y+1)=1$ hoặc $3$.
Nếu $\gcd=1$ thì $y+1=1$
$\Rightarrow y=0$
$\Rightarrow 3^x=1$
$\Rightarrow x=0$.
Nếu $\gcd=3$ thì $3\mid y+1$
$\Rightarrow y\equiv2\pmod3$
$\Rightarrow y^2-y+1\equiv4-2+1\equiv3\equiv0\pmod3$
Lại có $y^2-y+1=(y+1)^2-3y$ nên $9\nmid (y^2-y+1)$
Suy ra $y^2-y+1=3$
$\Rightarrow y^2-y-2=0$
$\Rightarrow y=2$.
Khi đó $3^x=2^3+1=9$ $\Rightarrow x=2$.
Vậy $\boxed{(x,y)=(0,0)\ \text{hoặc}\ (2,2).}$
b/
$a+b+c=0$
$\Rightarrow c=-(a+b)$
$ab+2c^2=ab+2(a+b)^2$
$=2a^2+5ab+2b^2$
$=(2a+b)(a+2b)$
Tương tự $bc+2a^2=(2a+b)(a-b)$
$ca+2b^2=(a+2b)(b-a)$
Suy ra $(ab+2c^2)(bc+2a^2)(ca+2b^2)$
$=(2a+b)^2(a+2b)^2(a-b)(b-a)$
$=-(2a+b)^2(a+2b)^2(a-b)^2$ $\le 0$
Do đó $N=1-(ab+2c^2)(bc+2a^2)(ca+2b^2)$$=1+(2a+b)^2(a+2b)^2(a-b)^2$
$\ge 1$$>0$
=> N là số dương.
1)
Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)
Dấu "=" xảy ra khi a=b=c
2)
\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)
Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)