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Vì \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ac}{abc}=0\Leftrightarrow ab+bc+ac=0\)
Ta có:
\(a+b+c=1\)
\(\Leftrightarrow\left(a+b+c\right)^2=1\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=1\)
\(\Leftrightarrow a^2+b^2+c^2=1\left(đpcm\right)\)
\(1.\)
Theo đề ra, ta có:
\(ax+by=c\)
\(bx+cy=a\Leftrightarrow ax+by+bx+cy+cx+ay=c+a+b\)
\(cx+by=b\)
\(\Leftrightarrow x\left(a+b+c\right)+y\left(a+b+c\right)=a+b+c\)
\(\Leftrightarrow\left(x+y-1\right)\left(a+b+c\right)=0\)
Ta có: \(x,y\)thỏa mãn \(\Rightarrow a+b+c=0\Rightarrow a+b=\left(-c\right)\)
Khi đó ta có:
\(a^3+b^3+c^3=a^3+3ab\left(a+b\right)+b^3-3ab\left(a+b\right)+c^3\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3=\left(-c\right)^3-3ab\left(-c\right)+c^3=3abc\)\(\left(đpcm\right)\)
\(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow a^2+b^2+c^2+3=2a+2b+2c\)
\(\Leftrightarrow a^2-2a+1+b^2-2b+1+c^2-2c+1=0\) \(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}a-1=0\\b-1=0\\c-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\Rightarrow a=b=c=1\Rightarrowđpcm\)
\(a^2+b^2+c^2+3=2\left(a+b+c\right)\Leftrightarrow a^2+b^2+c^2+3-2\left(a+b+c\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(a-1\right)^2\ge0\\\left(b-1\right)^2\ge0\\\left(c-1\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)
Dấu "=" xảy ra khi \(\left(a-1\right)^2=\left(b-1\right)^2=\left(c-1\right)^2=0\)
<=>a-1=b-1=c-1=0<=>a=b=c=1(đpcm)
2)
Xét hiệu:
\(A^2+B^2+C^2+D^2+4-2A-2B-2C-2D\)
\(=\left(A^2-2A+1\right)+\left(B^2-2B+1\right)+\left(C^2-2C+1\right)+\left(D^2-2D+1\right)\)
\(=\left(A-1\right)^2+\left(B-1\right)^2+\left(C-1\right)^2+\left(D-1\right)^2\ge0\)
=> BĐT luôn đúng
Vậy \(A^2+B^2+C^2+D^2+4\ge2\left(A+B+C+D\right)\)
1)
Áp dụng BĐT Cauchy cho 2 số không âm, ta có:
\(\dfrac{AB}{C}+\dfrac{BC}{A}\ge2\sqrt{\dfrac{AB}{C}.\dfrac{BC}{A}}=2B\) (1)
\(\dfrac{BC}{A}+\dfrac{AC}{B}\ge2\sqrt{\dfrac{BC}{A}.\dfrac{AC}{B}}=2C\) (2)
\(\dfrac{AB}{C}+\dfrac{AC}{B}\ge2\sqrt{\dfrac{AB}{C}.\dfrac{AC}{B}}=2A\) (3)
Từ (1)(2)(3) cộng vế theo vế:
\(2\left(\dfrac{AB}{C}+\dfrac{AC}{B}+\dfrac{BC}{A}\right)\ge2\left(A+B+C\right)\)
\(\Rightarrow\dfrac{AB}{C}+\dfrac{AC}{B}+\dfrac{BC}{A}\ge A+B+C\)
\(1,\)
Áp dụng BĐT Bunhiacopski:
\(A^2=\left(\sqrt{3-x}+\sqrt{x+7}\right)^2\le\left(1^2+1^2\right)\left(3-x+x+7\right)=2\cdot10=20\)
Dấu \("="\Leftrightarrow3-x=x+7\Leftrightarrow x=-2\)
\(A^2=3-x+x+7+2\sqrt{\left(3-x\right)\left(x+7\right)}\\ A^2=10+2\sqrt{\left(3-x\right)\left(x+7\right)}\ge10\)
Dấu \("="\Leftrightarrow\left(3-x\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)
1. Ta có : \(\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)
Tương tự : \(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\); \(\frac{1}{a^2}+\frac{1}{c^2}\ge\frac{2}{ac}\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\). Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=9\)
\(9\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c = 1
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=7\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=49\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}=49\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=49\)
biet ab+bc+ca=1CMR (a^2+1)(b^2+1)/(c^2+1)+(b^2+1)(c^2+1)/(a^2+1)+(a^2+1)(b^2+1)/(c^2+1) la binh phuong 1 so huu ti
\(VP-VT=\frac{\left(a-b\right)^2}{2b\left(a^2+b^2\right)}+\frac{\left(b-c\right)^2}{2a\left(b^2+c^2\right)}+\frac{\left(c-a\right)^2}{2b\left(c^2+a^2\right)}\)
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