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Áp dụng BĐT Cauchy schwarz dạng phân thức ta có :
\(\dfrac{a^2}{1+b-a}+\dfrac{b^2}{1+c-b}+\dfrac{c^2}{1+a-c}\ge\dfrac{\left(a+b+c\right)^2}{3}\ge\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\)
( vì \(a^2+b^2+c^2\ge ab+bc+ca\) )
Xảy ra đẳng thức khi và chỉ khi a=b=c= \(\sqrt{\dfrac{1}{3}}\)
Bài 2:
Ta có: \(f\left(a\right)=6a^5-10a^4-5a^3+23a^2-29a+2005\)
\(=\left(6a^5-10a^4-2a^3\right)-\left(3a^3-5a^2-a\right)+\left(18a^2-30a-6\right)+2011\)
\(=2a^3\left(3a^2-5a-1\right)-a\left(3a^2-5a-1\right)+6\left(3a^2-5a-1\right)+2011\)
\(=\left(2a^3-a+6\right)\left(3a^2-5a-1\right)+2011\)
Mà \(3a^2-5a-1=0\)
\(\Rightarrow f\left(a\right)=2011\)
Vậy...
Bạn chỉ cần để ý điều này thôi: \(\left(x-\frac{1}{x}\right)^2=x^2-2.x.\frac{1}{x}+\frac{1}{x^2}=x^2-2+\frac{1}{x^2}\)
Do đó giả thiết viết lại thành:
\(\left(a^2-2+\frac{1}{a^2}\right)+\left(b^2-2+\frac{1}{b^2}\right)+\left(c^2-2+\frac{1}{c^2}\right)=0\)
\(\Leftrightarrow\left(a-\frac{1}{a}\right)^2+\left(b-\frac{1}{b}\right)^2+\left(c-\frac{1}{c}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-\frac{1}{a}=0\\b-\frac{1}{b}=0\\c-\frac{1}{c}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{a}\\b=\frac{1}{b}\\c=\frac{1}{c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2=1\\b^2=1\\c^2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a^2\right)^{1010}=1^{1010}\\\left(b^2\right)^{1010}=1^{1010}\\\left(c^2\right)^{1010}=1^{1010}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^{2020}=1\\b^{2020}=1\\c^{2010}=1\end{matrix}\right.\) \(\Leftrightarrow a^{2020}+b^{2020}+c^{2020}=3\)
Sai thì bỏ qua ( bạn bè mà ) !
Nếu \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)
\(\Rightarrow\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=-1-1-1=-3\)(vô lí )
\(\Rightarrow a+b+c\ne0\)
Ta có :
\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=a+b+c\)
Đặt a + b + c = H
\(\Rightarrow\frac{a^2}{b+c}+\frac{ab}{a+c}+\frac{ac}{a+b}+\frac{b^2}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{c^2}{b+a}+\frac{ac}{c+b}+\frac{bc}{a+c}=H\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}+\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)+\left(\frac{ac}{a+b}+\frac{bc}{a+b}\right)+\left(\frac{ab}{b+c}+\frac{ac}{c+b}\right)=H\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}+a+b+c=H\)( Chỗ này làm hơi tắt bỏ qua nha )
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}=H-\left(a+b+c\right)\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}=0\left(đpcm\right)\)
ĐK:....
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a+b+c=a+b+c\)(nhân vào rồi tách)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
Việt Hoàng _ TTH (*Yonko Team*): Mình chưa xem kỹ nhưng có lẽ hướng làm bạn là sai òi nhé!
Ta có:
\(\sum\dfrac{ab+c}{c+1}=\sum\dfrac{ab+c}{a+c+b+c}\le\sum\dfrac{ab+c}{4}.\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)=\dfrac{a+b+c+3}{4}=\dfrac{4}{4}=1\)
Ta có : \(\frac{ab+c}{c+1}=\frac{ab+c\left(a+b+c\right)}{c+a+b+c}=\frac{a\left(b+c\right)+c\left(b+c\right)}{c+a+b+c}=\frac{\left(a+c\right)\left(b+c\right)}{c+a+b+c}\)
Do \(a;b;c>0\Rightarrow a+c;b+c>0\)
Áp dụng BĐT phụ : \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) , ta có :
\(\frac{ab+c}{c+1}\le\frac{\left(a+c\right)\left(b+c\right)}{4}\left(\frac{1}{c+a}+\frac{1}{b+c}\right)=\frac{\left(a+c\right)\left(b+c\right)}{4}.\frac{a+b+c+c}{\left(a+c\right)\left(b+c\right)}=\frac{c+1}{4}\left(1\right)\)
Tương tự , ta có : \(\frac{bc+a}{a+1}\le\frac{a+1}{4}\) ; \(\frac{ac+b}{b+1}\le\frac{b+1}{4}\left(2\right)\)
Từ ( 1 ) ; ( 2 ) có : \(\frac{ab+c}{c+1}+\frac{bc+a}{a+1}+\frac{ac+b}{b+1}\le\frac{a+1+b+1+c+1}{4}=\frac{a+b+c+3}{4}=1\)
Dấu " = " xảy ra <=> \(a=b=c=\frac{1}{3}\)
Vậy ...
Ta có:
\(\dfrac{1}{a+3b}+\dfrac{1}{c+3}\ge\dfrac{4}{a+3b+c+3}=\dfrac{4}{2b+6}=\dfrac{2}{b+3}\)
Tương tự:
\(\dfrac{1}{b+3c}+\dfrac{1}{a+3}\ge\dfrac{2}{c+3}\)
\(\dfrac{1}{c+3a}+\dfrac{1}{b+3}\ge\dfrac{2}{a+3}\)
Cộng vế:
\(\sum\dfrac{1}{a+3b}+\sum\dfrac{1}{a+3}\ge\sum\dfrac{2}{a+3}\)
\(\Rightarrow\sum\dfrac{1}{a+3b}\ge\sum\dfrac{1}{a+3}\) (đpcm)