Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{1}{ab}+2.\dfrac{1}{ac}+2.\dfrac{1}{bc}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\Leftrightarrow2\left(\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}\right)=0\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}=0\Leftrightarrow\dfrac{c+b+a}{abc}=0\Leftrightarrow a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow\left(a+b\right)^3=\left(-c\right)^3\Leftrightarrow a^3+b^3+3a^2b+3ab^2+c^3=0\Leftrightarrow a^3+b^3+c^3+3ab\left(a+b\right)=0\Leftrightarrow a^3+b^3+c^3-3abc=0\Leftrightarrow a^3+b^3+c^3=3abc\)
Vì \(3abc⋮3\)
Suy ra a3+b3+c3\(⋮3\)
\(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b+1}{12}+\dfrac{c+2}{18}\ge3\sqrt[3]{\dfrac{a^3\left(b+1\right)\left(c+2\right)}{216\left(b+1\right)\left(c+2\right)}}=\dfrac{a}{2}\)
Tương tự: \(\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c+1}{12}+\dfrac{a+2}{18}\ge\dfrac{b}{2}\)
\(\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}+\dfrac{a+1}{12}+\dfrac{b+2}{18}\ge\dfrac{c}{2}\)
Cộng vế:
\(VT+\dfrac{5}{36}\left(a+b+c\right)+\dfrac{7}{12}\ge\dfrac{1}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{13}{36}\left(a+b+c\right)-\dfrac{7}{12}\ge\dfrac{13}{36}.3\sqrt[3]{abc}-\dfrac{7}{12}=\dfrac{1}{2}\) (đpcm)
\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b+2}{36}+\dfrac{c+3}{48}\ge3\sqrt[3]{\dfrac{a^3\left(b+2\right)\left(c+3\right)}{1728\left(b+2\right)\left(c+3\right)}}=\dfrac{a}{4}\)
Tương tự: \(\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c+2}{36}+\dfrac{a+3}{48}\ge\dfrac{b}{4}\)
\(\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}+\dfrac{a+2}{36}+\dfrac{b+3}{48}\ge\dfrac{c}{4}\)
Cộng vế:
\(P+\dfrac{7\left(a+b+c\right)}{144}+\dfrac{17}{48}\ge\dfrac{a+b+c}{4}\)
\(\Rightarrow P\ge\dfrac{29}{144}\left(a+b+c\right)-\dfrac{17}{48}\ge\dfrac{29}{144}.3\sqrt[3]{abc}-\dfrac{17}{48}=\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
a) Từ giả thiết : \(\dfrac{1}{a}+\dfrac{1}{b}\text{=}\dfrac{1}{c}\)
\(\Rightarrow2ab\text{=}2bc+2ca\)
\(\Rightarrow2ab-2bc-2ca\text{=}0\)
Ta xét : \(\left(a+b-c\right)^2\text{=}a^2+b^2+c^2+2ab-2bc-2ca\)
\(\text{=}a^2+b^2+c^2\)
Do đó : \(A\text{=}\sqrt{a^2+b^2+c^2}\text{=}\sqrt{\left(a+b-c\right)^2}\)
\(\Rightarrow A\text{=}a+b-c\)
Vì a;b;c là các số hữu tỉ suy ra : đpcm
b) Đặt : \(a\text{=}\dfrac{1}{x-y};b\text{=}\dfrac{1}{y-x};c\text{=}\dfrac{1}{z-x}\)
Do đó : \(\dfrac{1}{a}+\dfrac{1}{b}\text{=}\dfrac{1}{c}\)
Ta có : \(B\text{=}\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\)
Từ đây ta thấy giống phần a nên :
\(B\text{=}a+b-c\)
\(B\text{=}\dfrac{1}{x-y}+\dfrac{1}{y-z}-\dfrac{1}{z-x}\)
Suy ra : đpcm.
Mình bổ sung đề phần b cần phải có điều kiện của x;y;z nha bạn.
Đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\c-a=z\end{matrix}\right.\Leftrightarrow x+y+z=0\)
\(\Leftrightarrow A=\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\\ \Leftrightarrow A=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)}\\ \Leftrightarrow A=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-\dfrac{2\left(x+y+z\right)}{xyz}}\\ \Leftrightarrow A=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-\dfrac{2\cdot0}{xyz}}\\ \Leftrightarrow A=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\left(đpcm\right)\)
3: \(\left\{{}\begin{matrix}a+b>=2\sqrt{ab}\\b+c>=2\sqrt{bc}\\a+c>=2\sqrt{ac}\end{matrix}\right.\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)>=8abc\)
1: =>(a+b)(a^2-ab+b^2)-ab(a+b)>=0
=>(a+b)(a^2-2ab+b^2)>=0
=>(a+b)(a-b)^2>=0(luôn đúng)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\)
Theo đề bài \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(\Rightarrow2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=0\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=0\)
\(\Rightarrow\frac{c+a+b}{abc}=0\) mà \(a;b;c\ne0\Rightarrow abc\ne0\Rightarrow a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Rightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)=-\left(a^3+b^3+c^3\right)\)
Mà \(3\left(a+b\right)\left(b+c\right)\left(c+a\right)\) chia hết cho 3 nên \(-\left(a^3+b^3+c^3\right)\) chia hết cho 3
Nên \(a^3+b^3+c^3\) chia hết cho 3