Cho \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)Chứng minh \(M=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)
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7 tháng 5 2017
Ta có: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1\)
=> \(\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right)=a+b+c\)
<=> \(\dfrac{a^2}{b+c}+\dfrac{ab}{b+c}+\dfrac{ac}{b+c}+\dfrac{b^2}{a+c}+\dfrac{ab}{a+c}+\dfrac{bc}{a+c}+\dfrac{c^2}{a+b}+\dfrac{ac}{a+b}+\dfrac{bc}{a+b}=a+b+c\)
<=> \(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+a\left(\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)+b\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)+c\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}\right)=a+b+c\)
<=> \(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+a+b+c=a+b+c\)
<=> \(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0\)
Vậy \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1\) thì \(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0\)
HP
24 tháng 2 2021
Hình như thế này mới đúng chứ \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\)
Áp dụng BĐT Cosi:
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2.\dfrac{a}{c};\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2.\dfrac{b}{a};\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2.\dfrac{c}{b}\)
\(\Rightarrow2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\)
Đẳng thức xảy ra khi \(a=b=c>0\)
21 tháng 3 2017
Bài 1:a,b,c ba cạnh tam giác => a,b,c dương
\(\left\{{}\begin{matrix}a+c>b\\a+b>c\\b+c>a\end{matrix}\right.\) ta có: \(\dfrac{x}{y}< \dfrac{x+p}{y+p}\forall_{x,y,p>0\&x< y}\)
\(VT=\dfrac{a}{a+b}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+c}{a+b}+\dfrac{b}{c+a}< \dfrac{a+c+c}{a+b+c}+\dfrac{b+b}{a+b+c}=\)
\(=\dfrac{a+b+c+b+c}{a+b+c}< \dfrac{\left(a+b+c\right)+\left(A+b+c\right)}{a+b+c}< \dfrac{2\left(b+a+c\right)}{a+b+c}=2=VP\)
p/s: đề sao làm vậy:
mình nghi đề phải thế này: \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\) cách làm đơn giản hơn
17 tháng 5 2018
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\) hả Lặng Thầm
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 3 2021
\(\Leftrightarrow\dfrac{2a^2}{b^2}+\dfrac{2b^2}{c^2}+\dfrac{2c^2}{a^2}=\dfrac{2a}{c}+\dfrac{2c}{b}+\dfrac{2b}{a}\)
\(\Leftrightarrow\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}-\dfrac{2a}{c}\right)+\left(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}-\dfrac{2c}{b}\right)+\left(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}-\dfrac{2b}{a}\right)=0\)
\(\Leftrightarrow\left(\dfrac{a}{b}-\dfrac{b}{c}\right)^2+\left(\dfrac{a}{b}-\dfrac{c}{a}\right)^2+\left(\dfrac{b}{c}-\dfrac{c}{a}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}-\dfrac{b}{c}=0\\\dfrac{a}{b}-\dfrac{c}{a}=0\\\dfrac{b}{c}-\dfrac{c}{a}=0\end{matrix}\right.\) \(\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Leftrightarrow a=b=c\)
18 tháng 6 2017
Áp dụng bất đẳng thức AM - GM ta có:
\(\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\le\dfrac{2}{2\sqrt{ab}}+\dfrac{2}{2\sqrt{bc}}+\dfrac{2}{2\sqrt{ac}}\)
\(=\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{bc}}+\dfrac{1}{\sqrt{ca}}\le\dfrac{1}{\sqrt{a^2}}+\dfrac{1}{\sqrt{b^2}}+\dfrac{1}{\sqrt{c^2}}\)
\(=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Dấu " = " xảy ra khi \(a=b=c\)
Vậy...
18 tháng 6 2017
Áp dụng BĐT \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\). Tương tự cho 2 BĐT còn lại có:
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c};\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{a+c}\)
Cộng theo vế 3 BĐT trên ta có:
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
Đẳng thức xảy ra khi \(a=b=c\)
8 tháng 9 2017
Ta có:
\(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1\)
=>\(\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right)=a+b+c\)
=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+c\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}\right)+b\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)+a\left(\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)=a+b+c\)
=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+a+b+c=a+b+c\)
=>\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)(đpcm)
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\Leftrightarrow\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c\)\(\Leftrightarrow a+\dfrac{a^2}{b+c}+b+\dfrac{b^2}{c+a}+c+\dfrac{c^2}{a+b}=a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\left(dpcm\right)\)